The answer is linear functions and downward-facing parabolas, which can easily be shown to work. Observe that equality always holds for linear \(f\), so we are free to shift \(f\) by any linear function while preserving the given condition; moreover, we can scale \(f\) by any positive constant. It is easy to rearrange the given equation into the following form: Claim: [Rewritten FE] For all \(x<y<z\), \[\frac{f(z)-f(x)}{z-x}-\frac{f(y)-f\big(\frac{x+z}2\big)}{y-\frac{x+z}2}\]is either 0 or has the same sign as \(y-\frac{x+z}2\). We will call the above assertion \(P(x,y,z)\). First, I contend: Claim: [Continuity and concavity] \(f\) is continuous at every point; moreover, for all \(x\), \(z\), we have \[f\left(\frac{x+z}2\right)\ge\frac{f(x)+f(z)}2.\] Proof. Assume by shifting that \(f(-1)=f(1)=0\). We will show that \(f(0)\ge0\), and that \(f\) is continuous at 0. First, applying \(P(-1,y,1)\) gives \(f(y)\le f(0)\) for all \(-1<y<1\). Applying \(P(-1,0,1-\varepsilon)\) for small \(\varepsilon>0\), we have \begin{align*} \frac{f(0)-f(-\varepsilon/2)}{\varepsilon/2}&\le\frac{f(1-\varepsilon)}{2-\varepsilon}\\ \implies f(0)&\le\frac{f(1-\varepsilon)}{4/\varepsilon-2}+f(-\varepsilon/2)\le\frac{f(0)}{4/\varepsilon-2}+f(-\varepsilon/2). \end{align*}Therefore, \[\left(1-\frac1{4/\varepsilon-2}\right)f(0)\le f(-\varepsilon/2)\le f(0),\]so \(f\) is continuous at \(0^-\), and since \(1-\frac1{4/\varepsilon-2}<1\), we also have \(f(0)\ge0\). Analogously, \(P(-1+\varepsilon,0,1)\) gives continuity at \(0^+\), so \(f\) is continuous at 0. \(\blacksquare\) Claim: [Derivative condition] \(f\) is differentiable at every point; moreover, \[f'(y)=\frac{f(y+a)-f(y-a)}{2a}\]for all \(y\), \(a\). Proof. Again, assume \(f(-1)=f(1)=0\). We will show \(f\) is differentiable at 0, and that \(f'(0)=0\). Limiting \(\varepsilon\to0^+\) in \(P(-1,0,1-\varepsilon)\) \[0\le\lim_{\varepsilon\to0^+}\frac{f(0)-f(-\varepsilon/2)}{\varepsilon/2}\le\lim_{\varepsilon\to0^+}\frac{f(1-\varepsilon)}{2-\varepsilon}=0,\]and similarly \(\lim_{\varepsilon\to0^+}\frac{f(\varepsilon/2)-f(0)}{\varepsilon/2}=0\), so \(f'(0)=0\). \(\blacksquare\) Now we will prove \(f\) quadratic; to this end, assume by shifting that \(f(0)=0\) and \(f'(0)=0\). The above claim implies that for all \(y\), \(a\), \(b\), \[\frac{f(y+a)-f(y-a)}{2a}=\frac{f(y+b)-f(y-b)}{2a}.\]We know that for all \(r\), \(f(r)=f(-r)\). We will proceed by strong induction on \(n\) to show that \(f(nr)=n^2f(r)\) for integers \(n\ge-1\). The base cases \(n=-1,0,1\) are handled already. To complete the inductive step, assume the hypothesis holds for integers less than \(n\). Then take \(y=n-2\), \(a=2\), \(b=1\) to derive \[\frac{f(nr)-(n-4)^2f(r)}4=\frac{(n-1)^2f(r)-(n-3)^2f(r)}2.\]It readily follows that \(f(nr)=n^2f(r)\). Finally, \(f(p/q)=p^2/q^2\cdot f(1)\) follows, so by continuity, \(f(x)\equiv x^2f(1)\). Since \(f\) is concave, \(f\) is either linear or a downward-facing parabola, so we are done.

The proof will proceed in seven steps. Prove that $f(x)=ax^2+bx+c$ for $a \leq 0$ works. Prove that $f\left(\frac{x+y}{2}\right) \geq \frac{f(x)+f(y)}{2}$ for all $x, y \in \mathbb R$. Prove that $f$ is "concave down"; that is, it always lies on or above its secant line. Prove that $f$'s left- and right-derivatives exist. Prove that $f$ has a derivative. Prove that $\frac{f(z)-f(w)}{z-w} = \frac{f(y)-f(x)}{y-x}$ whenever $w+z=x+y$. Obtain the solution set in Step 1. It helps to interpret the given inequality graphically. Indeed, the condition states that $f(y)$ lies "most above the secant line segment between $x$ and $z$" when $y = \frac{x+z}{2}$:

Let $P(x,y,z)$ be the given inequality. Suppose some function $f$ works. We proceed with a series of lemmas. Lemma: We have \[f\left(\frac{x+z}{2}\right)\ge\frac{f(x)+f(z)}{2}\]for all $x<z$. In particular, this shows that $P(x,y,z)$ holds for all $y\in[x,z]$. Proof: Suppose that $f(x)\le f(z)$ (the other case is similar), and suppose for the sake of contradiction that \[c:=\frac{f(x)+f(z)}{2} - f\left(\frac{x+z}{2}\right)>0.\]Let $A=(x,f(x)-c)$ and $B=(y,f(y)-c)$. We see that $P(x,y,z)$ implies that $(y,f(y))$ is below the line $AB$ for $x<y<z$. Suppose we pick $y\in(\tfrac{x+z}{2},\tfrac{1}{4}x+\tfrac{3}{4}z)$. Let $C=(2y-z,f(2y-z))$, $D=(y,f(y))$, and $E=(z,f(z))$. We see that $C$ and $D$ are below line $AB$, and $P(2y-z,y,z)$ implies that $F=(\tfrac{x+z}{2},f(\tfrac{x+z}{2}))$ is below the line through $D$ parallel to $CE$. [asy][asy] unitsize(10cm); pair A=(0,0); pair B=(1,0.5); pair F=(A+B)/2; pair E=B+0.2*dir(90); pair D=(0.7,0.26); pair C=(0.4,0.05); pair X=C-(C+E-2*D)/2; pair Y=E-(C+E-2*D)/2; draw(A--B); dot("$A$",A,dir(180)); dot("$B$",B,dir(0)); dot("$F$",F,dir(90)); dot("$E$",E,dir(90)); dot("$D$",D,dir(90)); dot("$C$",C,dir(90)); draw(1.2*X-0.2*Y--1.2*Y-0.2*X,dotted); [/asy][/asy] Note that $C$ is below $AB$ and $E$ is above $AB$, so the slope of $CD$ is greater than the slope of $AB$. Thus, $F$ is actually above the line through $D$ parallel to $CE$, which is the desired contradiction. This proves the lemma. $\blacksquare$ Let \[g(x):=\max\{|f(x-1)-f(x)|,|f(x+1)-f(x)|\}.\] Lemma: We have \[\left|\frac{f(x)-f(x-h)}{h}\right|\le g(x)\]for all $x$ and $|h|<1$. Proof: We prove the lemma in the case that $h>0$ and $f(x-h)<f(x)$ (the other cases are similar). Start by shifting $f$ such that $f(x)=0$, so let $f(x-h)=-N$ where $N>0$. We claim now that \[f(x-kh)\le -kN\]for all positive integers $k$. We prove this by induction on $k$, with $k=1$ given trivially. Now, suppose the claim is true for all $k'<k$. Firstly, if $k$ is even, then \[\frac{f(x-kh)+f(x)}{2}\le f(x-\tfrac{k}{2}h)\le -\tfrac{k}{2}N,\]which implies $f(x-kh)\le -kN$. If $k$ is odd, then \[\frac{f(x-kh)+f(x-h)}{2}\le f(x-\tfrac{k+1}{2}h)\le -\tfrac{k+1}{2}N,\]which implies $f(x-kh)\le -kN$. This complets the induction. Let $\ell$ be the line through $(x,0)$ with slope $N/h$. We see that $P(x-2t h,y,x)$ implies that $(y,f(y))$ is below $\ell$ for $y\in[x-2th,x-th]$ if $t$ is a positive integer (this is since $(x-th,f(x-th))$ is below $\ell$ and the slope between $(x-2th,f(x-2th))$ and $(x,0)$ is greater than $N/h$). Applying this where $t$ are powers of $2$, we actually see that $(y,f(y))$ is below $\ell$ for all $y\le x-h$. Thus, we see that $(x-1,f(x-1))$ is below $\ell$, so shifting back, \[\frac{f(x)-f(x-1)}{1}\ge \frac{N}{h},\]or \[g(x)\ge \left|\frac{f(x)-f(x-h)}{h}\right|,\]as desired. $\blacksquare$ This lemma actually provides some strong constraints on $f$. Lemma: We have that $f$ is continuous everywhere, and $f$ is differentiable almost everywhere (i.e. the set $S$ of $x$ where $f'(x)$ does not exist is a set of measure $0$). Proof: We first show that $f$ is continuous at $x$. If $g(x)=0$, then the first lemma actually implies that $f$ is constant from $x-1$ to $x+1$, so $f$ is trivially continuous at $x$. So suppose now that $g(x)>0$. Fix any $\varepsilon>0$. Note that if $|h|<\min\{1,\tfrac{\varepsilon}{g(x)}\}$, then \[|f(x-h)-f(x)|\le |h|\cdot g(x)\le \varepsilon.\]Thus, $f$ satisfies the standard definition of continuity by selecting $\delta=\min\{1,\tfrac{\varepsilon}{g(x)}\}$, so $f$ is continuous. Fix an integer $r$. Note that $[r,r+1]$ is compact, so $f$ achieves a maximum on $[r,r+1]$ as $f$ is continuous, so in fact $g$ reaches a maximum. Say $g(x)\le M$ for all $x\in[r,r+1]$. Thus, we see that \[|f(x)-f(y)|\le |x-y|\cdot M\]for $x,y\in[r,r+1]$ by the previous lemma. Thus, $f$ is Lipschitz on $[r,r+1]$, so $f$ is differentiable almost everywhere on this interval. Taking the union over all such intervals, we show that $f$ is differentiable almost everywhere. $\blacksquare$ Lemma: We have \[f'\left(\frac{x+z}{2}\right) = \frac{f(x)-f(z)}{x-z}\]whenever $\tfrac{x+z}{2}\not\in S$. Proof: This follows more or less trivially from $P(x,y,z)$, but we'll outline the details for completeness. Let $m=\tfrac{f(x)-f(z)}{x-z}$. We see that \[f'\left(\frac{x+z}{2}\right) = \lim_{y\to 0}\frac{f(\tfrac{x+z}{2})-f(y)}{\tfrac{x+z}{2}-y}.\]We see that $P(x,y,z)$ implies that $(y,f(y))$ is always below the line through $(\tfrac{x+z}{2},f(\tfrac{x+z}{2}))$ with slope $m$. Thus, the left limit of the above limit is at least $m$, and the right limit is at most $m$, so the limit must be exactly $m$, as desired. $\blacksquare$ We can now turn this into a useful functional equation to finally solve for $f$. Lemma: We have \[\frac{f(a)-f(b)}{a-b} = \frac{f(c)-f(a+b-c)}{2c-a-b}\]for all pairwise distinct real numbers $a,b,c$. Proof: Note that the lemma follows immediately if $\tfrac{a+b}{2}\not\in S$ from the differential equation. The key is now that $\mathbb{R}\setminus S$ is dense, so there exists arbitrarily small $\delta$ such that $\tfrac{a+b}{2}+\delta\not\in S$, so \[\frac{f(a+\delta)-f(b+\delta)}{a-b} = \frac{f(c+\delta)-f(a+b-c+\delta)}{2c-a-b}\]for arbitrarily small $\delta$. Since $f$ is continuous, we see that the above implies the desired FE. $\blacksquare$ We now solve the FE, with a method suggested by Ankit Bisain. Let $P$ be a quadratic polynomial such that $P(0)=f(0)$, $P(1)=f(1)$, and $P(\pi) = f(\pi)$. Call $x$ good if $f(x)=P(x)$. The above FE implies that if $a,b,c$ are distinct and good, then $a+b-c$ is also good. Therefore, $m+n\pi$ is good for all integers $m,n$, so \[f(m+\pi n)=P(m+\pi n).\]The set $\{m+\pi n:m,n\in\mathbb{Z}\}$ is dense in $\mathbb{R}$, and $f$ is continuous, so we see that $f(x)=P(x)$ for all real $x$. It is easy to see that the FE is satisfied if and only if $P''(0)\le 0$. Thus, the answer to the problem is $\boxed{f(x)=ax^2+bx+c}$, where $a,b,c\in\mathbb{R}$ and $a\le 0$.

Here's the official solution. Answer: all functions of the form $f(y) = a y^2 + by + c$, where $a, b, c$ are constants with $a \leq 0$. If $I = (x,z)$ is an interval, we say that a real number $\alpha$ is a supergradient of $f$ at $y \in I$ if we always have \[ f(t) \le f(y) + \alpha(t-y) \]for every $t \in I$. (This inequality may be familiar as the so-called ``tangent line trick''. A cartoon of this situation is drawn below for intuition.) We will also say $\alpha$ is a supergradient of $f$ at $y$, without reference to the interval, if $\alpha$ is a supergradient of some open interval containing $y$. [asy][asy] size(4cm); pair X = (0,0); pair Z = (7,3); pair Y = (3,4); draw(X..Y..Z, blue); pair t = 3*dir(X..Y..Z, 1); draw( (Y-t)--(Y+t), red, Arrows); label(rotate(15)*"slope $\alpha$", Y+0.45*t, dir(90), red ); dot("$x$", X, dir(-90), blue); dot("$z$", Z, dir(-90), blue); dot("$y$", Y, dir(90), blue); [/asy][/asy] \qquad [asy][asy] size(5cm); pair X = (-3,-9/3); pair Z = (2,-4/3); draw( X..(-2,-4/3)..(-1,-1/3)..(0,0)..(1,-1/3)..Z, blue ); pair Y = (-0.5,-0.25/3); draw(X--Z, deepgreen); draw((Y-1.5*dir(Z-X))--(Y+1.5*dir(Z-X)), red, Arrows ); label(rotate(18)*"slope $\frac{f(z)-f(x)}{z-x}$", midpoint(X--Z), dir(110), deepgreen); dot("$x$", X, dir(-90), blue); dot("$y = \frac{x+z}{2}$", Y, 1.5*dir(-60), blue); dot("$z$", Z, dir(-90), blue); [/asy][/asy] Claim: The problem condition is equivalent to asserting that $\frac{f(z) - f(x)}{z-x}$ is a supergradient of $f$ at $\frac{x+z}{2}$ for the interval $(x,z)$, for every $x < z$. Proof. This is just manipulation. $\blacksquare$ At this point, we may readily verify that all claimed quadratic functions $f(x) = ax^2+bx+c$ work --- these functions are concave, so the graphs lie below the tangent line at any point. Given $x < z$, the tangent at $\frac{x+z}{2}$ has slope given by the derivative $f'(x)=2ax+b$, that is \[ f'\left(\frac{x+z}{2}\right) = 2a \cdot \frac{x+z}{2} + b = \frac{f(z)-f(x)}{z-x} \]as claimed. (Of course, it is also easy to verify the condition directly by elementary means, since it is just a polynomial inequality.) Now suppose $f$ satisfies the required condition; we prove that it has the above form. Claim: The function $f$ is concave. Proof. Choose any $\Delta > \max\{z-y,y-x\}$. Since $f$ has a supergradient $\alpha$ at $y$ over the interval $(y-\Delta,y+\Delta)$, and this interval includes $x$ and $z$, we have \begin{align*} \frac{z-y}{z-x}f(x) + \frac{y-x}{z-x}f(z) &\leq \frac{z-y}{z-x}(f(y) + \alpha(x-y)) + \frac{y-x}{z-x}(f(y) + \alpha(z-y)) \\ &= f(y). \end{align*}That is, $f$ is a concave function. Continuity follows from the fact that any concave function on ${\mathbb R}$ is automatically continuous. $\blacksquare$ Lemma: [see e.g. {https://math.stackexchange.com/a/615161} for picture] Any concave function $f$ on ${\mathbb R}$ is continuous. Proof. Suppose we wish to prove continuity at $p \in {\mathbb R}$. Choose any real numbers $a$ and $b$ with $a < p < b$. For any $0 < \varepsilon < \max(b-p,p-a)$ we always have \[ f(p) + \frac{f(b)-f(p)}{b-p} \varepsilon \le f(p+\varepsilon) \le f(p) + \frac{f(p)-f(a)}{p-a} \varepsilon \]which implies right continuity; the proof for left continuity is the same. $\blacksquare$ Claim: The function $f$ cannot have more than one supergradient at any given point. Proof. Fix $y \in {\mathbb R}$. For $t > 0$, let's define the function \[ g(t) = \frac{f(y)-f(y-t)}{t} - \frac{f(y+t)-f(y)}{t}. \]We contend that $g(\varepsilon) \le \frac35g(3\varepsilon)$ for any $\varepsilon > 0$. Indeed by the problem condition, \noindent \begin{minipage}{0.5\textwidth} \begin{align*} f(y) &\le f(y-\varepsilon) + \frac{f(y+\varepsilon)-f(y-3\varepsilon)}{4} \\ f(y) &\le f(y+\varepsilon) - \frac{f(y+3\varepsilon)-f(y-\varepsilon)}{4}. \end{align*}Summing gives the desired conclusion. \end{minipage} \begin{minipage}{0.4\textwidth} [asy][asy] size(6cm); real f(real x) { return -x*x/6; } pair A = (-3, f(3)); pair B = (-2, f(2)); pair C = (-1, f(1)); pair D = (-0, f(0)); pair E = (1, f(1)); pair F = (2, f(2)); pair G = (3, f(3)); draw(A..B..C..D..E..F..G, blue); draw(A--E, lightred); draw(C--G, deepgreen); draw( (C-2*dir(E-A))--(C+2*dir(E-A)), lightred, Arrows ); draw( (E-2*dir(G-C))--(E+2*dir(G-C)), deepgreen, Arrows ); dot("$y-3\varepsilon$", A, dir(-90), blue); dot("$y-\varepsilon$", C, dir(-90), blue); dot("$y$", D, dir(-90), blue); dot("$y+\varepsilon$", E, dir(-90), blue); dot("$y+3\varepsilon$", G, dir(-90), blue); [/asy][/asy] Now suppose that $f$ has two supergradients $\alpha < \alpha'$ at point $y$. For small enough $\varepsilon$, we should have we have $f(y-\varepsilon) \leq f(y) - \alpha'\varepsilon$ and $f(y+\varepsilon) \leq f(y) + \alpha\varepsilon$, hence \[ g(\varepsilon) = \frac{f(y)-f(y-\varepsilon)}{\varepsilon} - \frac{f(y+\varepsilon)-f(y)}{\varepsilon} \geq \alpha' - \alpha. \]This is impossible since $g(\varepsilon)$ may be arbitrarily small. $\blacksquare$ Claim: The function $f$ is quadratic on the rational numbers. Proof. Consider any four-term arithmetic progression $x, x+d, x+2d, x+3d$. Because $(f(x+2d)-f(x+d))/d$ and $(f(x+3d)-f(x))/3d$ are both supergradients of $f$ at the point $x+3d/2$, they must be equal, hence \[ f(x+3d) - 3f(x+2d) + 3f(x+d) - f(x) = 0. \]If we fix $d = 1/n$, it follows inductively that $f$ agrees with a quadratic function $\widetilde f_n$ on the set $\frac1n {\mathbb Z}$. On the other hand, for any $m \neq n$, we apparently have $\widetilde f_n = \widetilde f_{mn} = \widetilde f_m$, so the quadratic functions on each ``layer'' are all equal. $\blacksquare$ Since $f$ is continuous, it follows $f$ is quadratic, as needed. Remark: [Alternate finish using differentiability due to Michael Ren] In the proof of the main claim (about uniqueness of supergradients), we can actually notice the two terms $\frac{f(y)-f(y-t)}{t}$ and $\frac{f(y+t)-f(y)}{t}$ in the definition of $g(t)$ are both monotonic (by concavity). Since we supplied a proof that $\lim_{t \to 0} g(t) = 0$, we find $f$ is differentiable. Now, if the derivative at some point exists, it must coincide with all the supergradients; (informally, this is why ``tangent line trick'' always has the slope as the derivative, and formally, we use the mean value theorem). In other words, we must have \[ f(x+y) - f(x-y) = 2f'(x) \cdot y \]holds for all real numbers $x$ and $y$. By choosing $y=1$ we obtain that $f'(x) = f(x+1) - f(x-1)$ which means $f'$ is also continuous. Finally differentiating both sides with respect to $y$ gives \[ f'(x+y) - f'(x-y) = 2f'(x) \]which means $f'$ obeys Jensen's functional equation. Since $f'$ is continuous, this means $f'$ is linear. Thus $f$ is quadratic, as needed.

The solutions are all polynomials of degree $\leq 2$ with non-positive second degree coefficient, which obviously work. Call the inequality $P(x,y,z)$ and notice that linear functions always give equality, so if $f$ is a solution $f+l$ is a solution as well, where $l$ is any linear function. This allows us to define for each $u<v$ the function $f_{u,v}=f-l$, where $l$ is the linear function for which $l(u)=f(u)$ and $l(v)=f(v)$, so that $f_{u,v}(u)=f_{u,v}(v)=0$, and $f_{u,v}$ satisfies the inequality. Also, call $P_{u,v}$ the corresponding statement. CLAIM 1: $f_{u,v}$ is non-decreasing on $(u,\frac{u+v}{2})$ and non-increasing on $(\frac{u+v}{2},v)$. First observe that $P_{u,v}(u,t,v)$ gives $f_{u,v}(t) \leq f_{u,v}(\frac{u+v}{2})$ for all $t \in (u,v)$. Then consider any $a,b \in (u,\frac{u+v}{2})$ with $a<b$ and pick any $c \in (u+v-b,u+v-a)$, and consider $P_{u,v}(a,\frac{u+v}{2},c)$ and $P_{u,v}(b,\frac{u+v}{2},c)$. Using $f_{u,v}(\frac{a+c}{2}) \leq f_{u,v}(\frac{u+v}{2})$ and $f_{u,v}(\frac{b+c}{2}) \leq f_{u,v}(\frac{u+v}{2})$, with some manipulation the former gives $f_{u,v}(a) \leq f_{u,v}(c)$, while the latter gives $f_{u,v}(b) \geq f_{u,v}(c)$, so $f_{u,v}(a) \leq f_{u,v}(b)$, as we wanted. The proof for $a,b \in (\frac{u+v}{2},v)$ is completely analogous. $\square$ CLAIM 2: $f_{u,v}(t) \geq 0$ for all $t \in (u,v)$. Suppose there exists $k \in (u,\frac{u+v}{2})$ with $f_{u,v}(k) < 0$, and pick $m \in (\frac{k}{2}, k)$. Non-decreasingness gives $f_{u,v}(\frac{k}{2}) \leq f_{u,v}(m) \leq f_{u,v}(k)$, but $P_{u,v}(u,m,k)$ rearranges to $f_{u,v}(m) \leq f_{u,v}(\frac{k}{2})+\frac{2m-u-k}{2(k-u)}f_{u,v}(k)$. Since the fraction is positive, we get $f_{u,v}(m) < f_{u,v}(\frac{k}{2})$, contradiction. So $f_{u,v}$ is non-negative on $(u,\frac{u+v}{2})$, and the proof for $(\frac{u+v}{2},v)$ is analogous. Since $f_{u,v}(t) \leq f_{u,v}(\frac{u+v}{2})$ for all $t \in (u,v)$, we have $f_{u,v}(\frac{u+v}{2}) \geq 0$ as well. $\square$ CLAIM 3: $f$ is concave and continuous. Concavity follows from CLAIM 2, and it is well-known that concave $\implies$ continuous. $\square$ CLAIM 4: there exists a function $h(x)$ such that $h(x)=\frac{f(x+a)-f(x-a)}{2a}$ for all $a>0$. Choose $x,p,q$ with $0<p<q$. Notice that to show $\frac{f(x+p)-f(x-p)}{2p}=\frac{f(x+q)-f(x-q)}{2q}$ it is sufficient to show $f_{x-q,x+q}(x-p)=f_{x-q,x+q}(x+p)$. For $t \in (x,x+q)$ combining $P_{x-q,x+q}(x-q,t,x+q)$ and $P_{x-q,x+q}(x-p,x,t)$ gives $f_{x-q,x+q}(t) \geq f_{x-q,x+q}(x-p)$ for $t<x+p$ and $f_{x-q,x+q}(t) \leq f_{x-q,x+q}(x-p)$ for $t>x+p$, so by continuity it must be the case that $f_{x-q,x+q}(x-p)=f_{x-q,x+q}(x+p)$, which is what we needed. $\square$ CLAIM 5: $f$ is differentiable, and its derivative is equal to $h$. For $\epsilon \in (0,1)$ we have $\frac{f(x+\epsilon)-f(x)}{\epsilon}=h(x+\frac{\epsilon}{2})=\frac{f(x+1+2\epsilon)-f(x-1)}{2+2\epsilon}$. The expression on the right is continuous and has a limit of $h(x)$ as $\epsilon$ approaches $0^+$, so the expression on the right has the same limit. Analogously we can show the left limit exists and is equal to $h(x)$ as well, so the claim follows. $\square$ CLAIM 6: the only solutions to the functional equation $g'(\frac{x+y}{2})=\frac{g(x)-g(y)}{x-y}$ for all $x \neq y$ are quadratic polynomials. Since quadratic polynomials satisfy the equation, we can shift $g$ by an appropriate polynomial to get $\hat g$, such that $\hat g(0)= \hat g (2) = \hat g (2\sqrt{2})=0$. Then obviously $\hat g'(1)=\hat g'(\sqrt{2})=\hat g'(1+\sqrt{2})=0$. This implies $\hat g(1+x)=\hat g(1-x)$, $\hat g(\sqrt{2}+x)=\hat g(\sqrt{2}-x)$ and $\hat g(1+\sqrt{2}+x)=\hat g(1+\sqrt{2}-x)$. Combining the first and the third gives $\hat g'$ has period $2\sqrt{2}$, while the second and the third give a period of $2$. So $\hat g'(2m+2\sqrt{2}n)$ for all $m,n \in \mathbb{Z}$, and the set of the values of $2m+2\sqrt{2}n$ is dense in $\mathbb{R}$, we have that $\hat g$ is zero everywhere by continuity, so $g$ was a quadratic polynomial at the beginning. $\square$ CLAIM 6 gives that $f$ is a quadratic polynomial, and concavity says the leading term is non-positive, so we get the solutions are what we said at the beginning. $\blacksquare$

v_Enhance wrote: Find all functions $f \colon \mathbb{R} \to \mathbb{R}$ that satisfy the inequality \[ f(y) - \left(\frac{z-y}{z-x} f(x) + \frac{y-x}{z-x}f(z)\right) \leq f\left(\frac{x+z}{2}\right) - \frac{f(x)+f(z)}{2} \]for all real numbers $x < y < z$. Proposed by Gabriel Carroll Excellent problem! This problem is one of a kind. To begin with, let \(P(x,y,z)\) assert the given functional inequality. Notice that for any linear function \(g\), if \(f\) is a solution, so is \(f+g\). Therefore, we are free to fix any two values. Taking this to our advantage, I fix \(f(1)=f(-1)=0\) (note this is temporary). I now prove a claim. Claim 1. \(f\) is concave and continuous. Proof. I divide this into two parts. The first part is proving \(f\) continuous. Let \(\epsilon\) be an extremely small positive real. \(P(-1,y,1)\) gives us that \(f(y)\leq f(0)\) for all \(y\in (-1,1)\). \(P(-1,0,1-\epsilon)\) gives us that \[f(0)-f\left(-\frac{\epsilon}{2}\right)\leq\frac{\epsilon f(1-\epsilon)}{2(2-\epsilon)}\]so \[f(0)\geq f\left(-\frac{\epsilon}{2}\right)\geq f(0)-\frac{\epsilon f(1-\epsilon)}{2(2-\epsilon)}\]and as \(\epsilon\to0\), we see that \(f\left(-\frac{\epsilon}{2}\right)=f(0)\). Therefore, \(f\) is continuous at \(0^-\). Similarly, \(P(-1+\epsilon,0,1)\) tells us that \(f\) is continuous at \(0^+\), implying that \(f\) is continuous at \(0\). Since we can always shift \(f\) by a linear function, and all linears are continuous, we conclude that \(f\) is continuous at all points. Also, note that \(f(0)\geq 0\). This implies that when \(a,b\) satisfy \(f(a)=f(b)=0\), then \(f\left(\frac{a+b}{2}\right)\geq0\) (use the same logic). Now, assume for the sake of contradiction, that there exist \(a,b\) such that \(f\) is not concave. Shift \(f\) so that \(f(a)=f(b)=0\). Then, you get \(0> f\left(\frac{a+b}{2}\right)\), a contradiction. Therefore \(f\) is concave and continuous as claimed. $\blacksquare$ Claim 2. \(f\) is differentiable at all points. Proof. \(P(-1,0,1-\epsilon)\) gives us that \[0\leq f'(0)=\lim_{\epsilon\to 0+}\frac{f(0)-f\left(-\frac{\epsilon}{2}\right)}{\frac{\epsilon}{2}}\leq\lim_{\epsilon\to 0^+}\frac{f(1-\epsilon)}{2-\epsilon}=0\]implying that \(f'(0)=0\) and that \(f\) is differentiable at all points. Again, shifting tells us that \(f\) is differentiable at all points, as claimed. $\blacksquare$ Claim 3. \(f'\left(\frac{x+y}{2}\right)=\frac{f(x)-f(y)}{x-y}\). Proof. Firstly, note that this fact holds when \(f(x)=f(y)=0\). So, if you scale \(f\) up by a linear function, the LHS of the equation adds the leading coefficient of the linear and the RHS adds the same. Since the fact holds true for \(f(x)=f(y)=0\), it also holds for any other images in \(f\), since we can scale it up by a linear function, as claimed. $\blacksquare$ Finally, back to the problem, we see that \[\frac{f(x+a)-f(x-a)}{2a}=\frac{f(x+b)-f(x-b)}{2b}\]for all reals \(x,a,b\) from claim 3, call this \(Q(x,a,b)\). Now shift \(f\) such that \(f'(0)=f(0)=0\). Then \(f\) is even. Then \(Q(a,a,b)\) gives us that \[\frac{b}{a}f(2a)=f(a+b)-f(a-b)\]Switching \(a\) and \(b\) and using the fact that \(f\) is even gives us that \[\frac{a}{b}f(2b)=\frac{b}{a}f(2a)\]for all non-zero \(a,b\). If \(f(a)=0\) for one \(a\), then \(f(x)=0\) for all \(x\) because of the equation we got above, so let all images be non-zero. Then, \[\frac{f(2a)}{f(2b)}=\frac{(2a)^2}{(2b)^2}\]implying that \(\frac{f(x)}{x^2}\) is constant. Therefore \(f(x)=cx^2\). Note that \(c\) is negative, because \(f\) is concave. Freeing the conditions \(f(0)=0\) and \(f'(0)=0\), we see that the solutions are all linears and all concave parabolas, which indeed fit!

I would like to wholeheartedly thank CANBANKAN for nudging me

Let $P(x,y,z)$ denote the given assertion. We claim that all functions that work are $f(x)=mx^2+nx+k$ with $m$ non-positive. Lemma 1: $f$ is continuous and concave. Proof. WLOG by shifting $f(1)=f(-1)=0.$ We have $f(0)\geq f(x)$ for $x\in (-1,1).$ Pick a small $k\in (1,0)$ then $P(k-1,0,1)$ and $P(-1,0,1-k)$ respectively imply $f$ continuous at $0^+$ an $0^-$. We also have $f(0)$ is non-negative. $\blacksquare$ Lemma 2: $f$ is differentiable. Proof. WLOG by shifting $f(1)=f(-1)=0.$ As $k$ approaches $0^-$ and $0^+$ by $P(k-1,0,1)$ and $P(-1,0,1-k)$ we get $\lim_{k\to 0^-} \tfrac{f(k/2)-f(0)}{k/2} =0$ and $\lim_{k\to 0^+} \tfrac{f(0)-f(-k/2)}{k/2}=0$ implying the claim. $\blacksquare$ In particular we have $2f'(x)=\tfrac{f(x+y)-f(x-y)}{y}.$ Form which we easily get $f'$ is continuous. Now differentiating in respect to $y$ gives $f'(x+y)-f'(x-y)=2f'(x).$ This is Jensen. Q.E.D.

First, notice $f(x)$ is concave down trivially at any point $a$ by $(a - \infty, a, a + \infty)$. Next, notice $f(x)$ is continuous at any point $a$ by using the concave down condition and looking at points $a-1, a+1$. Next, notice $f(x)$ is differentiable at any point $a$ by using the continuous condition on $(a - 1, a, a+1)$ and $(a-1+\epsilon, a +\epsilon, a +1+\epsilon)$ for infinitesimal epsilon, both positive and negative cases. Next, notice that $\frac{f(y)-f(x)}{y-x}$ depends solely on the value of $y+x$ by differentiability, since there is only one value of $f'(x)$. This alone is enough to conclude that $f(x) = ax^2+bx+c$ where $a \leq 0$ over the rational numbers by first solving over the integers and using rational gcd. Now since $f$ is continuous, the same conclusion holds for all real numbers, so we are done.

What a terrible problem. I claim that the solution set is $f(x)=ax^2+bx+c$, where $a<0$. Let $a=\frac{x+z}{2}$, let $b=\frac{z-x}{2}$, and let $c=y-a$. We can rewrite the given FE to say $$f(a+c)+\frac{cf(a-b)}{2b}-\frac{cf(a+b)}{2b} \le f(a)$$for all reals $a$, $b$, $c$ subject to the constraint $b>c>-b$. Note that $f$ is invariant when shifted by a linear function and scaling by a positive constant, so WLOG $f(0)=0$ and $f(1)=f(-1)$, and scale so $|f(1)|=1$. Consider the equations formed by substituting $(a,b,c)$ and $(a,b,-c)$. Add these to get $f(a+c)+f(a-c) \le 2f(a)$. Putting $(a,c)=(0,1)$ into this implies that $f(1)=f(-1)=-1$. Now consider $g(x)=f(x)+x^2$. Subsituting $f(x)=g(x)-x^2$ into our FE yields $$g(a+c)+\frac{cg(a-b)}{2b}-\frac{cg(a+b)}{2b} \le g(a) +c^2$$. Now we have $g(0)=g(1)=g(-1)$. Rearranging this yields $$g(a+c)-g(a) \le c^2 + c(\frac{g(a-b)-g(a+b)}{2b})$$. Subsitute $(a,b,c)=(a+nc,c,b)$ to get $$g(a+(n+1)c)-g(a+nc) \le c^2 + c\frac{g(a-b)-g(a+b)}{2b}$$. Summing this as $n$ varies from $0$ to $k-1$ yields $$g(a+ck)-g(a) \le kc^2+ck\frac{g(a-b)-g(a+b)}{2b}$$. Fixing $ck=d$ and taking the limit as $c$ approaches $0$ yields $$g(a+d)-g(a) \le d\frac{g(a-b)-g(a+b)}{2b}$$. We can again do the trick where we subsitute both $(a,b,d)$ and $(a,b,-d)$ to get $2g(a) \ge g(a+d)+g(a-d)$. I now claim that this concavity equation implies that if three numbers $x$, $y$, $z$ are in arithmetic progression in that order, and $f(x)=0$, $f(y)=0$, and $f(z)=0$, then $f(\frac{x+y}2)=0$ and $f(\frac{y+z}2)=0$. To do this is suffices to check this is true for $(x,y,z)=(-1,0,1)$. To do this, take $(a,d)=(\frac12,\frac12)$ and $(a,d)=(-\frac12,-\frac12)$ yields $g(\frac12) \ge 0$ and $g(-\frac12) \ge 0$. Now substituting $(a,d)=(0,\frac12)$ into our concavity equation yields the desired result. Thus, because we can take midpoints, we find that $g(x)=0$ for all rational numbers $x$ between $-1$ and $1$ which have terminating decimals in base $2$. Importantly, these form a dense set. Because the rate of increase of $g$ is bounded from above, and because $g$ has a dense set of roots, this implies $g(x)$ is identically $0$ on the interval $[-1,1]$. All that remains is to extend this to all of the real numbers. To do this, I check by induction that $g(x)$ is identically $0$ on the range $[-1,k]$ for positive integers $k$. The proof for $[-k,1]$ is analogous. Suppose we have proved $g(x)$ is $0$ on the range $[-1,k]$. Then by the pseudo-concavity we know that $g$ is nonpositive on $[k,k+1]$. Now in the most recent equation I wrote with $a$, $b$, and $d$, pick $(a,b,d)=(k-0.1,0.1+m,0.05)$ with $m \le 1$ to get that $g$ is nonnegative on $[k,k+1]$. Thus by induction $g$ is identically $0$ on the real numbers, so we are done.

I conjectured a year ago that this problem would be property spam. I'm going to say that yeah it kinda just is property spam. Some of the above solutions seem slightly fishy. We claim that all quadratic functions with negative leading coefficient and linear functions work. It is easy to check this holds. Note that this is equivalent to saying that if the segment from $(x, f(x))$ to $(z, f(z))$ is moved vertically such that $\left(\frac{x+z}{2}, f\left(\frac{x+z}{2}\right)\right)$ lies on it, all other points on the line lie below it. As such, it follows that $f\left(\frac{x+z}{2}\right) \ge \frac{f(x) + f(z)}{2}$. Claim: For any $x < y < z$, the point $(y, f(y))$ lies above or on the line from $(x, f(x))$ to $(z, f(z))$. Proof. WLOG scale such that $x = f(x) = 0$ and $z = f(z) = 1$. Now, if $y = 2^{-a} \cdot n$ for integers $a, n$, this follows immediately. Else, suppose that $y$ is not of that form and lies below the line. Then it follows that one of $2y, 2y - 1$ lies in $(0, 1)$ lies twice as far below the line. Repeating this until $f(y) < -10^{1434}$ and $|y - 0.5| > \frac{1}{8}$. Applying the claim with $a$ and the element of $\{0, 1\}$ it is farther from implies that $f(0.5) < 0.5$, contradiction. $\blacksquare$ Claim: If points $(x, f(x)), (y, f(y)), (z, f(z))$ are ever collinear then $f$ is a line. Proof. WLOG let $x < y < z$. If any element on the line from $(x, f(x))$ to $(z, f(z))$ lies above it, then taking that element and one of $x$ or $z$ gives a contradiction. Thus, $f$ is linear on the interval $[x, z]$. We can again use this claim on $x' \in \left[\frac32 x - \frac12 z, x\right]$, $\frac{x' + z'}{2}, z' \in [x,z]$ to get that all $x'$ are also linear in the same manner. Repeat this to get the entire function is linear. $\blacksquare$ Claim: $f$ is continuous. Proof. Fix an arbitrary point $x$. Let the slope from $f(x - c)$ to $f(x + c)$ be $m$ for fixed $c$. Then all elements of the graph in a small neighborhood of $(x, f(x))$ lie below the line with slope $m$ through $(x, f(x))$. Now, take a $d$ such that $f(x - d) \ne f(x)$, which is evidently possible or else $f$ is linear. Then it follows that for all small $\varepsilon$, $(x - \varepsilon, f(x - \varepsilon))$ is stomached between two lines that approach $x$, which finishes by squeeze theorem. $\blacksquare$ Claim: $f$ is left-differentiable (and thus right-differentiable). Proof. We have similarly that the slope from $(x - \varepsilon, f(x - \varepsilon))$ is monotonic and bounded, which finishes. $\blacksquare$ Claim: $f$ is differentiable. Proof. FTSOC suppose that at some point $x$, it has left derivative with slope $e_1$ and right derivative with slope $e_2$ which are unequal. Take small enough neighborhood such that all slopes to $x$ are within $\delta < (e_1 - e_1) \cdot 10^{-1434}$ of their actual value. $f$ is then effectively piecewise. However, then taking $x - 3\varepsilon$ and $x + \varepsilon$ gives a contradiction in slope. $\blacksquare$ It thus follows that for a fixed $x$, the slope of $\frac{f(x + c) - f(x - c)}{2c}$ is fixed. WLOG subtract a linear function such that $f$ is not injective, from which it follows that is even around some point. Then by Lagrange interpolation, we can subtract a polynomial $P$ to get $g = f - P$ such that $g$ has $3$ zeros, from which it follows that $g$ is zero.