Never mind

Alternatively, after obtaining $f(a^3)+f(b^3)+f(c^3)=3f(abc)\quad \forall a+b+c=0$, we can take $a+b+c+d=0$ and use it in the given equation in two ways, first as $(a+b) +c+d=0$ and then as $a+(b+c) +d=0$, compare the two equations thus obtained and conclude that $f$ satisfied Jensen's functional equation. Combined with the increasing nature of $f$, this implies that $f$ is linear.

Wrong

We claim that all solutions are of the form $f(x)\equiv cx+d$ for some $c\ge 0$. It's easy to check all such functions work, because of the identity $a^3+b^3+c^3-3abc=\tfrac12(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2).$ If $f$ is a function satisfying the conditions, then so is $f+k$ for any constant $k$, so we can assume WLOG $f(0)=0$. Note that the given statements imply: $``$For reals $a,b,c$ with $a+b+c=0$, we have $f(a^3)+f(b^3)+f(c^3)=3f(abc)."$ Call this statement $P(a,b,c)$. Now $P(a^{\frac13},-a^{\frac13},0)$ gives $f(a)+f(-a)=0$, so $f$ is odd. Now for $a>0$, we have $a^{\frac13}+0+0\ge 0$, so condition $(i)$ gives $f(a)\ge 0$. Because of oddness, we have $f(a)\le 0$ for $a<0$. So $f$ takes non-negative values at non-negative inputs and non-positive values for non-positive inputs. Now consider that statement $P(a,b,-(a+b))$. Because $f$ is odd, this translates into$$f(a^3)+f(b^3)+3f(ab(a+b))=f((a+b)^3).$$Using this equation multiple times, we obtain \begin{align*} f((a+b+c)^3) &=f((a+b)^3)+f(c^3)+3f((a+b)c(a+b+c))\\ &= f(a^3)+f(b^3)+3f(ab(a+b))+f(c^3)\\ &+3f((a+b)c(a+b+c))\end{align*}Similarly, switching the roles of $b$ and $c$ we get, $$f((a+b+c)^3)=f(a^3)+f(b^3)+f(c^3)+3f(ac(a+c))+3f((b(a+c)(a+b+c)).$$Comparing these two expressions for $f((a+b+c)^3)$, we obtain $$f(c(a+b)(a+b+c))+f(ab(a+b))=f(b(a+c)(a+b+c))+f(ac(a+c)).\qquad (\star)$$Now choose arbitrary positive reals $x\le y$ and consider the following system of equations \begin{align} a^2c-abc&=a^2b+ab^2\\ a^2b+ab^2&=x\\ a^2c+ac^2&=y \end{align}We claim that this has a solution in reals $a,b,c$. Indeed, set $b=qa,c=ra$, and $(1)$ becomes $$r-qr=q+q^2\implies r=\frac{q^2+q}{1-q}.$$And $(2)\div (3)$ becomes $$\frac{x}{y}=\frac{q(q+1)}{r(r+1)}=\frac{q(q+1)}{\frac{q(q+1)}{1-q}\cdot\frac{q^2+1}{1-q}}=\frac{(q-1)^2}{1+q^2}.$$The function $h(q)=\tfrac{(q-1)^2}{1+q^2}$ is continuous on $[0,1]$ and $h(0)=1,h(1)=0$. Since $\tfrac xy\in(0,1]$, there is a $q\in[0,1)$ so that $h(q)=\tfrac xy$. Choose this $q$, and the corresponding $r=\tfrac{q(q+1)}{1-q}$. Then scale both $q$ and $r$ by $a$ to get $b,c$ so that $(2)$ holds; then $(3)$ and $(1)$ would hold automatically. Thus $a,b,c$ exist; clearly $a\ne 0$. Now that we've chosen suitable $a,b,c$, note that $$a^2c-abc=a^2b+ab^2\implies b(a+b+c)=ac\implies b(a+c)(a+b+c)=ac(a+c)=y,$$and $$ab(a+b)=x,$$So $$c(a+b)(a+b+c)=(b(a+c)(a+b+c)+ac(a+c))-ab(a+b)=2y-x.$$Using these in $(\star)$ gives $$f(2y-x)+f(x)=2f(y)\; \forall 0<x\le y.$$Setting $z=2y-x\iff y=\frac{x+z}{2}$, this becomes $$f(x)+f(z)=2f\left(\frac{x+z}{2}\right)\;\forall 0<x\le z.$$So $f$ satisfies Jensen's functional equation over positives, and it's bounded on $[0,\infty)$, so $f(x)=cx+d$ for some $c\ge 0$, as claimed. $\blacksquare$

I just proceeded slightly. This was my approach. We guess the solution to be $f(x) = cx + a$ for $c\geq 0$ and $a,c \in \mathbb{R}$. Let $g(x) = f(x) - a$. So our equation for $x+y+z \geq$ becomes $g(x) + g(y) + g(z) - 3a = f(x) + f(y) + f(z) \geq 3f(xyz) = g(xyz) - 3a$. So, $g(x) + g(y) + g(z) = 3g(xyz)$. We get a similar equation for $x+y+z \leq 0$. So, We can shift our function by any constant. So, WLOG $f(0) = 0$. Plugging $a,-a,0$ for some positive $a$, we get $f(a^3) + f(-a^3) + 0 = 0$. So, we get $f$ is odd. Let $a^3>b^3$ then plugging $a, -b, 0$, we get $f(a^3) - f(b^3) + 0 = 0$. $f(a^3) > f(b^3)$. So, $f$ is increasing.

@ankoganit what is jensen's functional equation do you have a link.I searched it on google but couldn't find anything clear.

The equation $f(x)+f(y)=2f\left(\frac{x+y}{2}\right)$ is often called Jensen's functional equation. I can't find an online resource right now, but here's a proof that Jensen on $\mathbb R^+$ implies linear.

Ankoganit wrote: The equation $f(x)+f(y)=2f\left(\frac{x+y}{2}\right)$ is often called Jensen's functional equation. I can't find an online resource right now, but here's a proof that Jensen on $\mathbb R^+$ implies linear. Or you can take $f \rightarrow f+c$, because the equation is invariant under affine transformation, and you end up with cauchy after taking $f(0)=0$.

HERE IS MINE:JUSIFY HOW MUCH I CAN GET OUT OFF 7

Attachments:
soln elmo-day 2-prob 4.pdf (24kb)

Here's the official solution from the ELMO shortlist: The answer is $f(x) = kx + \ell$ where $k$ and $\ell$ are any real numbers with $k \ge 0$. Since $f$ can be shifted by a constant, WLOG $f(0) = 0$. $c = 0, b = -a$ gives that $f(-a)=-f(a)$. Then plugging in $c=0$ and using this, we get that $f$ is increasing. Now, we let $c = -a - b$ to get: \[f(a^3)+f(b^3)+f(-(a+b)^3) = 3f(-ab(a+b))\]Using $-f(x)=-f(x)$ and rearranging: \[f(a^3)+f(b^3) + 3f(ab(a+b)) = f((a+b)^3)\]Call this property $P(a, b)$. Lemma: $f(2^k m) = 2^k f(m)$ for all integer $k$ and real $m > 0$. Proof. $P(d^{1/3}, d^{1/3})$ gives $2f(d) + 3f(2d) = f(8d)$. Consider the sequence $\alpha_k = f(2^km)$. We have a linear recurrence: $\alpha_{k+3} = 3\alpha_{k+1} + 3\alpha_k$. Its characteristic equation has roots $2, -1, -1$, so we have $f(2^km) = \alpha_k = c_12^k + c_2(-1)^k+c_3(-1)^kk$ for some $c_1, c_2, c_3$ that may depend on $m$ but not on $k$. This can be extended to negative $k$ as well. Note that since $f(x)$ is increasing and $f(0) = 0$, $\alpha_k \ge 0$ for all $k$. Now, if either $c_2$ or $c_3$ is nonzero, you can take $k \rightarrow -\infty$ with the right parity, and you will get $\alpha_k < 0$, a contradiction. Thus $c_2=c_3=0$, so $f(2^km) = c_12^k$. Plugging in $k=0$, we get $c_1 = f(m)$, so $f(2^km) = 2^kf(m)$ as desired. $\blacksquare$ Lemma: $f(\phi^{3k}m) = \phi^{3k}f(m)$ for all integer $k$ and real $m > 0$. Proof. $P(d^{1/3}, \phi d^{1/3})$ gives $f(d) + 4f(\phi^3d) = f(\phi^6d)$. Again, this gives a linear recurrence for the sequence $\beta_k = \phi^{3k}m$, $\beta_{k+2} = 4\beta_{k+1} + \beta_k$. Its characteristic equation has roots $\phi^3, -\phi^{-3}$, so we have $f(\phi^{3k}m) = \beta_k = c_4\phi^{3k} + c_5(-\phi^{-3})^k$ for some $c_4, c_5$ that may depend on $m$ but not on $k$. As before, $c_5$ must be zero, so $f(\phi^{3k}m) = c_4\phi^{3k}$. Plugging in $k=0$, $c_4=f(m)$, so $f(\phi^{3k}m) = \phi^{3k}f(m)$ as desired. $\blacksquare$ Now I claim that $f(x) = f(1)x$ for all $x$. Since $f(-x) = -f(x)$, we only need to prove this for positive $x$. If $f(1) = 0$, we are done by Lemma 1. Otherwise, for a contradiction, let $f(n) \neq f(1)n$ for some $n>0$. (note that $f(n) \ge 0$). Let $f(n) > f(1)n$; the case where $f(n) < f(1)n$ is similar. By Dirichlet's approximation theorem, we can find $r, s$ such that: \[n < \frac{2^s}{\phi^{3r}} < \frac{f(n)}{f(1)}\]or, expanding, \[\phi^{3r}n < 2^s \implies \phi^{3r}f(n) > 2^sf(1)\]But, by Lemmas 1 and 2: \[ f(\phi^{3r}n) = \phi^{3r}f(n) \qquad\text{and}\qquad f(2^s) = 2^sf(1)\]a contradiction to the fact that $f$ is increasing. Thus, $f(x) = f(1)x$ for all $x$. Re-adjusting for the assumption that $f(0) = 0$, $f(x)$ is linear. Plugging back in to the condition, $f(x)$ can be any linear function with a nonnegative coefficient of $x$.

Again, my solution is the same as official one. It took me 4.5hr. By the way, I can't send pictures through my phone so I didn't took part in the test. Also the picture size is limited by 512KB. But I do share my solution with IMO 2018 China team leader.

There is another finish after $f(2x)=2f(x)$: without too much effort, we can generalise it to $f(nx)=nf(x)$ by plugging $(x,(n-1)x,-nx)$ and applying the same "linear recurrence" trick. Then we are done, since we have solved it for all rationals.

v_Enhance wrote: Here's the official solution from the ELMO shortlist: The answer is $f(x) = kx + \ell$ where $k$ and $\ell$ are any real numbers with $k \ge 0$. Since $f$ can be shifted by a constant, WLOG $f(0) = 0$. $c = 0, b = -a$ gives that $f(-a)=-f(a)$. Then plugging in $c=0$ and using this, we get that $f$ is increasing. Now, we let $c = -a - b$ to get: \[f(a^3)+f(b^3)+f(-(a+b)^3) = 3f(-ab(a+b))\]Using $-f(x)=-f(x)$ and rearranging: \[f(a^3)+f(b^3) + 3f(ab(a+b)) = f((a+b)^3)\]Call this property $P(a, b)$. Lemma: $f(2^k m) = 2^k f(m)$ for all integer $k$ and real $m > 0$. Proof. $P(d^{1/3}, d^{1/3})$ gives $2f(d) + 3f(2d) = f(8d)$. Consider the sequence $\alpha_k = f(2^km)$. We have a linear recurrence: $\alpha_{k+3} = 3\alpha_{k+1} + 3\alpha_k$. Its characteristic equation has roots $2, -1, -1$, so we have $f(2^km) = \alpha_k = c_12^k + c_2(-1)^k+c_3(-1)^kk$ for some $c_1, c_2, c_3$ that may depend on $m$ but not on $k$. This can be extended to negative $k$ as well. Note that since $f(x)$ is increasing and $f(0) = 0$, $\alpha_k \ge 0$ for all $k$. Now, if either $c_2$ or $c_3$ is nonzero, you can take $k \rightarrow -\infty$ with the right parity, and you will get $\alpha_k < 0$, a contradiction. Thus $c_2=c_3=0$, so $f(2^km) = c_12^k$. Plugging in $k=0$, we get $c_1 = f(m)$, so $f(2^km) = 2^kf(m)$ as desired. $\blacksquare$ Lemma: $f(\phi^{3k}m) = \phi^{3k}f(m)$ for all integer $k$ and real $m > 0$. Proof. $P(d^{1/3}, \phi d^{1/3})$ gives $f(d) + 4f(\phi^3d) = f(\phi^6d)$. Again, this gives a linear recurrence for the sequence $\beta_k = \phi^{3k}m$, $\beta_{k+2} = 4\beta_{k+1} + \beta_k$. Its characteristic equation has roots $\phi^3, -\phi^{-3}$, so we have $f(\phi^{3k}m) = \beta_k = c_4\phi^{3k} + c_5(-\phi^{-3})^k$ for some $c_4, c_5$ that may depend on $m$ but not on $k$. As before, $c_5$ must be zero, so $f(\phi^{3k}m) = c_4\phi^{3k}$. Plugging in $k=0$, $c_4=f(m)$, so $f(\phi^{3k}m) = \phi^{3k}f(m)$ as desired. $\blacksquare$ Now I claim that $f(x) = f(1)x$ for all $x$. Since $f(-x) = -f(x)$, we only need to prove this for positive $x$. If $f(1) = 0$, we are done by Lemma 1. Otherwise, for a contradiction, let $f(n) \neq f(1)n$ for some $n>0$. (note that $f(n) \ge 0$). Let $f(n) > f(1)n$; the case where $f(n) < f(1)n$ is similar. By Dirichlet's approximation theorem, we can find $r, s$ such that: \[n < \frac{2^s}{\phi^{3r}} < \frac{f(n)}{f(1)}\]or, expanding, \[\phi^{3r}n < 2^s \implies \phi^{3r}f(n) > 2^sf(1)\]But, by Lemmas 1 and 2: \[ f(\phi^{3r}n) = \phi^{3r}f(n) \qquad\text{and}\qquad f(2^s) = 2^sf(1)\]a contradiction to the fact that $f$ is increasing. Thus, $f(x) = f(1)x$ for all $x$. Re-adjusting for the assumption that $f(0) = 0$, $f(x)$ is linear. Plugging back in to the condition, $f(x)$ can be any linear function with a nonnegative coefficient of $x$. Great Solution! Anyway, it's not so trivial when f(1)=0.

v_Enhance wrote: Here's the official solution from the ELMO shortlist: The answer is $f(x) = kx + \ell$ where $k$ and $\ell$ are any real numbers with $k \ge 0$. Since $f$ can be shifted by a constant, WLOG $f(0) = 0$. $c = 0, b = -a$ gives that $f(-a)=-f(a)$. Then plugging in $c=0$ and using this, we get that $f$ is increasing. Now, we let $c = -a - b$ to get: \[f(a^3)+f(b^3)+f(-(a+b)^3) = 3f(-ab(a+b))\]Using $-f(x)=-f(x)$ and rearranging: \[f(a^3)+f(b^3) + 3f(ab(a+b)) = f((a+b)^3)\]Call this property $P(a, b)$. Lemma: $f(2^k m) = 2^k f(m)$ for all integer $k$ and real $m > 0$. Proof. $P(d^{1/3}, d^{1/3})$ gives $2f(d) + 3f(2d) = f(8d)$. Consider the sequence $\alpha_k = f(2^km)$. We have a linear recurrence: $\alpha_{k+3} = 3\alpha_{k+1} + 3\alpha_k$. Its characteristic equation has roots $2, -1, -1$, so we have $f(2^km) = \alpha_k = c_12^k + c_2(-1)^k+c_3(-1)^kk$ for some $c_1, c_2, c_3$ that may depend on $m$ but not on $k$. This can be extended to negative $k$ as well. Note that since $f(x)$ is increasing and $f(0) = 0$, $\alpha_k \ge 0$ for all $k$. Now, if either $c_2$ or $c_3$ is nonzero, you can take $k \rightarrow -\infty$ with the right parity, and you will get $\alpha_k < 0$, a contradiction. Thus $c_2=c_3=0$, so $f(2^km) = c_12^k$. Plugging in $k=0$, we get $c_1 = f(m)$, so $f(2^km) = 2^kf(m)$ as desired. $\blacksquare$ Lemma: $f(\phi^{3k}m) = \phi^{3k}f(m)$ for all integer $k$ and real $m > 0$. Proof. $P(d^{1/3}, \phi d^{1/3})$ gives $f(d) + 4f(\phi^3d) = f(\phi^6d)$. Again, this gives a linear recurrence for the sequence $\beta_k = \phi^{3k}m$, $\beta_{k+2} = 4\beta_{k+1} + \beta_k$. Its characteristic equation has roots $\phi^3, -\phi^{-3}$, so we have $f(\phi^{3k}m) = \beta_k = c_4\phi^{3k} + c_5(-\phi^{-3})^k$ for some $c_4, c_5$ that may depend on $m$ but not on $k$. As before, $c_5$ must be zero, so $f(\phi^{3k}m) = c_4\phi^{3k}$. Plugging in $k=0$, $c_4=f(m)$, so $f(\phi^{3k}m) = \phi^{3k}f(m)$ as desired. $\blacksquare$ Now I claim that $f(x) = f(1)x$ for all $x$. Since $f(-x) = -f(x)$, we only need to prove this for positive $x$. If $f(1) = 0$, we are done by Lemma 1. Otherwise, for a contradiction, let $f(n) \neq f(1)n$ for some $n>0$. (note that $f(n) \ge 0$). Let $f(n) > f(1)n$; the case where $f(n) < f(1)n$ is similar. By Dirichlet's approximation theorem, we can find $r, s$ such that: \[n < \frac{2^s}{\phi^{3r}} < \frac{f(n)}{f(1)}\]or, expanding, \[\phi^{3r}n < 2^s \implies \phi^{3r}f(n) > 2^sf(1)\]But, by Lemmas 1 and 2: \[ f(\phi^{3r}n) = \phi^{3r}f(n) \qquad\text{and}\qquad f(2^s) = 2^sf(1)\]a contradiction to the fact that $f$ is increasing. Thus, $f(x) = f(1)x$ for all $x$. Re-adjusting for the assumption that $f(0) = 0$, $f(x)$ is linear. Plugging back in to the condition, $f(x)$ can be any linear function with a nonnegative coefficient of $x$. Excuse me, can you explain how does the Dirichlet's approximation theorem work?

I would like to know how do you use this approximation theorem, dear Evan, as to me it's very vague. If anybody wondered what's $\phi$ (as it took me about 15 minutes to figure out) this is $\frac{1+\sqrt5}{2}$

liekkas wrote: v_Enhance wrote: Here's the official solution from the ELMO shortlist: The answer is $f(x) = kx + \ell$ where $k$ and $\ell$ are any real numbers with $k \ge 0$. Since $f$ can be shifted by a constant, WLOG $f(0) = 0$. $c = 0, b = -a$ gives that $f(-a)=-f(a)$. Then plugging in $c=0$ and using this, we get that $f$ is increasing. Now, we let $c = -a - b$ to get: \[f(a^3)+f(b^3)+f(-(a+b)^3) = 3f(-ab(a+b))\]Using $-f(x)=-f(x)$ and rearranging: \[f(a^3)+f(b^3) + 3f(ab(a+b)) = f((a+b)^3)\]Call this property $P(a, b)$. Lemma: $f(2^k m) = 2^k f(m)$ for all integer $k$ and real $m > 0$. Proof. $P(d^{1/3}, d^{1/3})$ gives $2f(d) + 3f(2d) = f(8d)$. Consider the sequence $\alpha_k = f(2^km)$. We have a linear recurrence: $\alpha_{k+3} = 3\alpha_{k+1} + 3\alpha_k$. Its characteristic equation has roots $2, -1, -1$, so we have $f(2^km) = \alpha_k = c_12^k + c_2(-1)^k+c_3(-1)^kk$ for some $c_1, c_2, c_3$ that may depend on $m$ but not on $k$. This can be extended to negative $k$ as well. Note that since $f(x)$ is increasing and $f(0) = 0$, $\alpha_k \ge 0$ for all $k$. Now, if either $c_2$ or $c_3$ is nonzero, you can take $k \rightarrow -\infty$ with the right parity, and you will get $\alpha_k < 0$, a contradiction. Thus $c_2=c_3=0$, so $f(2^km) = c_12^k$. Plugging in $k=0$, we get $c_1 = f(m)$, so $f(2^km) = 2^kf(m)$ as desired. $\blacksquare$ Lemma: $f(\phi^{3k}m) = \phi^{3k}f(m)$ for all integer $k$ and real $m > 0$. Proof. $P(d^{1/3}, \phi d^{1/3})$ gives $f(d) + 4f(\phi^3d) = f(\phi^6d)$. Again, this gives a linear recurrence for the sequence $\beta_k = \phi^{3k}m$, $\beta_{k+2} = 4\beta_{k+1} + \beta_k$. Its characteristic equation has roots $\phi^3, -\phi^{-3}$, so we have $f(\phi^{3k}m) = \beta_k = c_4\phi^{3k} + c_5(-\phi^{-3})^k$ for some $c_4, c_5$ that may depend on $m$ but not on $k$. As before, $c_5$ must be zero, so $f(\phi^{3k}m) = c_4\phi^{3k}$. Plugging in $k=0$, $c_4=f(m)$, so $f(\phi^{3k}m) = \phi^{3k}f(m)$ as desired. $\blacksquare$ Now I claim that $f(x) = f(1)x$ for all $x$. Since $f(-x) = -f(x)$, we only need to prove this for positive $x$. If $f(1) = 0$, we are done by Lemma 1. Otherwise, for a contradiction, let $f(n) \neq f(1)n$ for some $n>0$. (note that $f(n) \ge 0$). Let $f(n) > f(1)n$; the case where $f(n) < f(1)n$ is similar. By Dirichlet's approximation theorem, we can find $r, s$ such that: \[n < \frac{2^s}{\phi^{3r}} < \frac{f(n)}{f(1)}\]or, expanding, \[\phi^{3r}n < 2^s \implies \phi^{3r}f(n) > 2^sf(1)\]But, by Lemmas 1 and 2: \[ f(\phi^{3r}n) = \phi^{3r}f(n) \qquad\text{and}\qquad f(2^s) = 2^sf(1)\]a contradiction to the fact that $f$ is increasing. Thus, $f(x) = f(1)x$ for all $x$. Re-adjusting for the assumption that $f(0) = 0$, $f(x)$ is linear. Plugging back in to the condition, $f(x)$ can be any linear function with a nonnegative coefficient of $x$. Great Solution! Anyway, it's not so trivial when f(1)=0. It is trivial since f is increasing.

Very Nice Problem We claim that only linear functions satisfy the conditions . It is easy to check that they work . We now prove that they are the only ones. First note that if $f(x)$ is a solution , so id $cf(x)+d$ where $c,d \in \mathbb{R}$ . So WLOG , we assume $f(0)=0$ and $f(1)=1$ . Next, note that $$a+b+c=0 \iff f(a^3)+f(b^3)+f(c^3) = 3f(abc)$$It turns out this identity is alone sufficient to solve the problem almost completely (along with the aid of brutal 3-variable bashing ofcourse ) . We call this as $P(a,b,c)$. Also note that $ P(a,-a,0) \implies \text {f is odd}.$ Further , the first condition implies that $a>0 \implies f(a) \geq 0$ . So now we restrict our attention to positive reals only . Let $a,b,c \in \mathbb{R^+}$ and let $s=a+b+c$ . Comparing $P(a+b,c,-s)$ and $P(a,b+c,-s)$ , we have $$f(sc(a+b))+f(ab(a+b))=f(sa(b+c))+f(bc(b+c))$$. Also, note that $sc(a+b)+ab(a+b)=sa(b+c)+bc(b+c)$ Call the last line $\spadesuit$ Now we present the following Lemma : For any arbitary reals $x,y \in \mathbb{R^+}$ with $x > y $ , there exists a triple $\{a,b,c\}$ of positive reals such that the following system of equations has a solution . \begin{align} a(a+b+c)&=bc(b+c)\\ c(a+b)(a+b+c)&=x\\ ab(a+b)&=y \end{align}Let $b=ka$ and $c=la$ with $k,l \in \mathbb{R^+}$ . $(1)$ yields $l= \frac {k+1}{k-1}$ . Now we note that $$\frac xy = \frac {l(1+k)(1+l+k)}{k(1+k)}= \frac { (\frac{k+1}{k-1})\cdot (\frac {k^2+k}{k-1})}{k}= (\frac {k+1}{k-1})^2 = l^2 $$ Next,note that the function $g(t)=t^2$ is continuos throughout. So we can guaraantee a choice of $l$ which also fixes our choice of $k$ . Finally choose $a$ as required . (as it is justa matter of scaling) . Also note that $h(t) = \frac {k+1}{k-1}$ can't take 1 , so we chose $x > y$ . Hence , from $(\spadesuit)$ we come to the conclusion that $f(x)+f(y)= 2 f(\frac {x+y}{2})$ for all $x,y \in \mathbb{R^+}$ . (We just proved it for $x \neq y$ and $x=y$ is trivial) . Hence , since $f$ satisfies Jensen's FE on positive reals and is bounded on $\mathbb {R^+}$, hence along with $f(0)=0$ and $f(1)=1$ , we conclude that $f(x)=x$ , $\forall x \in \mathbb{R^+}$ . But $f$ is odd so we have $\boxed {f(x)=x \text{ } \forall x \in \mathbb{R}}$ Shifting back from $f(0)=0$ and $f(1)=1$ , we conclude that all linear functions work.

ELMO 2017 P6 wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all real numbers $a,b,$ and $c$: If $a + b + c \ge 0$ then $f(a^3) + f(b^3) + f(c^3) \ge 3f(abc)$. If $a + b + c \le 0$ then $f(a^3) + f(b^3) + f(c^3) \le 3f(abc)$. Nice problem First, some housekeeping. The answers are all the functions of the form $\boxed{f(x) = mx + n}$ for any real constant $m \ge 0$. To prove this, notice that if $f(x) = mx + n, m \ge 0$, then \begin{align*} f(a^3) + f(b^3) + f(c^3) - 3f(abc) &= m(a^3 + b^3 + c^3 - abc) \\ &= \frac{1}{2} m (a + b + c)((a - b)^2 + (b - c)^2 + (c - a)^2) \end{align*}and since $m \ge 0$, this is nonnegative when $a + b + c \ge 0$, and nonpositive if $a +b + c \le 0$, which satisfies the condition of the problem. Let $P(a,b,c)$ be the assertion of $a,b,c$ to the "functional condition/inequality" above. Note that if $f$ is a solution, then $f + k$ is a solution as well. So we could WLOG $f(0) = 0$. Claim 01. $f$ is increasing. Proof. Consider $P(a,-b,0)$ where $a \ge b$. Then, we have $f(a) \ge f(b)$. Now, fix $a + b + c = 0$. Since this choice of $(a,b,c)$ satisfies both condition of the problem, then we must have equality, indeed \[ f(a^3) + f(b^3) + f(c^3) = 3f(abc) \]Claim 02. $f$ is odd. Proof. Notice that $P(a^{\frac{1}{3}},-a^{\frac{1}{3}},0)$ gives us $f(-a) = -f(a)$. Now, rewrite the condition as \begin{align*} f(a^3) + f(b^3) + f(-(a + b)^3) &= 3f(-ab(a + b)) \\ f(a^3) + f(b^3) + 3f(ab(a + b)) &= f((a + b)^3) \end{align*}since $f$ is odd. Now, name this assertion as $Q(a,b)$ instead. Claim 03. $f(\phi^{3i} \ell) = \phi^{3i}f(\ell)$ for all $i \in \mathbb{Z}$ and $\ell \in \mathbb{R}^+$. Proof. Let $\phi = \frac{\sqrt{5} + 1}{2}$ be the golden ratio. In particular, $\phi^2 = \phi + 1$. $Q(\phi^{i + 1} \ell^{\frac{1}{3}}, \phi^i \ell^{\frac{1}{3}})$ gives us \[ 4f(\phi^{3i + 3} \ell) + f(\phi^{3i} \ell) = f(\phi^{3i + 6} \ell) \]Let $f(\phi^{3j} \ell) = a_j$. Then, we have the following recursive equation: \[ a_{i + 2} = a_{i} + 4a_{i + 1} \]Solving this, we have $a_n = a(2 + \sqrt{5})^n + b(2 - \sqrt{5})^n$ for all $n \in \mathbb{Z}$. Suppose $b \not= 0$. Since we have $0 < |2 - \sqrt{5}| < 1$, then \begin{align*} \lim_{n \to -\infty} a_n &= \lim_{n \to -\infty} a(2 + \sqrt{5})^n + b(2 - \sqrt{5})^n \\ &< 0 \\ &= f(0) \\ &\le \lim_{n \to -\infty} f(\phi^{3n} \ell) \\ &= \lim_{n \to -\infty} a_n \end{align*}which is a contradiction as $f$ is increasing. So, $b = 0$. Hence, we conclude that $f(\phi^{3j} \ell) = a_j = a \cdot (2 + \sqrt{5})^j = a \cdot \phi^{3j}$ for a constant $a$. Substituting $j = 0$ gives $f(\ell) = a$. Hence, we conclude that $f(\phi^{3j} \ell) = \phi^{3j}f(\ell)$, as desired. Claim 04. $f(2^{i} \ell) = 2^{i}f(\ell)$ for all $i \in \mathbb{Z}$ and $\ell \in \mathbb{R}^+$. Proof. $P((2^i\ell)^{\frac{1}{3}}, (2^i \ell)^{\frac{1}{3}})$ gives us \[ 2f(2^{i} \ell) + 3f(2^{i + 1} \ell) = f(2^{i + 3} \ell) \]Let $f(2^j \ell) = b_j$. Then, we have the following recursive equation: \[ b_{i + 3} = 3b_{i + 1} + 2b_i \]Solving this, we have $b_n = a(2)^n + b(-1)^n + c(-1)^n \cdot n$ for all $n \in \mathbb{Z}$. Suppose $b,c \not= 0$. Then, notice that \begin{align*} \lim_{n \to -\infty} b_n &= \lim_{n \to -\infty} (a \cdot 2^n + b \cdot (-1)^n + cn \cdot (-1)^n) \\ &= \lim_{n \to -\infty} a \cdot 2^n + b \cdot (-1)^n - |c|n \\ &< 0 \\ &= f(0) \\ &\le \lim_{n \to -\infty} f(2^n \ell) \\ &= \lim_{n \to -\infty} b_n \end{align*}which is a contradiction as $f$ is increasing. So, $b,c=0$. Hence, we conclude that $f(2^j \ell) = b_j = a \cdot 2^j$ for a constant $a$. Substituting $j = 0$ gives us $f(\ell) = a$. Hence, we conclude that $f(2^j \ell) = 2^j f(\ell)$, as desired. Since $f$ is odd, it suffices to prove that $f(n) = nf(1)$ for all $n \in \mathbb{R}^+$. If $f(1) = 0$, then $f(2^j) = 0$ for all large integer $\ell$ and by the condition of $f$ is increasing, we have $f \equiv 0$. Otherwise, $f(1) > 0$. Suppose there exists $n \in \mathbb{R}^+$ such that $f(n) > nf(1)$. By Dirichlet Approximation, there exists $a,b \in \mathbb{Z}$ such that \[ \frac{f(n)}{f(1)} > \frac{2^a}{\phi^{3b}} > n \]Since $f$ is increasing, we have \[ f(n) < f \left( \frac{2^a}{\phi^{3b}} \right) = \frac{1}{\phi^{3b}}f(2^a) = \frac{2^a}{\phi^{3b}}f(1) \Leftrightarrow \frac{f(n)}{f(1)} < \frac{2^a}{\phi^{3b}} < \frac{f(n)}{f(1)}\]which is a contradiction. The other direction is similar. Therefore, $f(n) = nf(1)$ for all $n \in \mathbb{R}^+$ and since $f$ is odd, $f(n) = nf(1)$ for all $n \in \mathbb{R}$. Motivational Remark. The part before Claim 03 is quite straightforward for most people. However, Claim 03 is by no mean obvious. This is mainly inspired by \[ Q(a,1): f(\textcolor{blue}{a^3}) + f(1) + 3f(\textcolor{red}{a(a + 1)}) = f((a + 1)^3) \]Now, we want to pick a suitable real value $a$ such that we could have a lot of same terms: in this case, blue and red terms seem like a suitable choice: $a^3 = a(a + 1)$, motivating the plug $a = \phi$. Furthermore, this gives us something like \[ 4 f(\phi^{3i + 3}) + f(\phi^{3i}) = f(\phi^{3i + 6}) \]This is a recurrence relation. Furthermore, you could see that everything in the function is "homogeneous": in the sense that if you change all insertions $a \mapsto a t^{\frac{1}{3}}$, then we could get a similar identity, but everything inside $f$ is multiplied with any real number $t$. In fact, this assertion gives us $f(a_i t) = a_i f(t)$ for any real number $t$ and $a_i$, a sequence of constant. We now attempt to find something similar: $f(b_i t) = b_i f(t)$ for any real number $t$ and a sequence of constant $b_i$, and dense over $a_i$ in $\mathbb{R}$. Lastly, to prove that $f(n) = nf(1)$ for all $n \in \mathbb{R}^+$. The idea is to prove that $f(n) < nf(1)$ if $f(n) > nf(1)$ (similarly for the other direction), by taking some reals of the form $\frac{2^a}{(\phi^3)^b}$ in the interval $\left( n, \frac{f(n)}{f(1)} \right)$, which is possible since $\frac{2^a}{(\phi^3)^b}$ is "dense in $\mathbb{R}$." A more rigorous approach is possible by the well known Dirichlet Approximation.

Woke up and found the worse version of intended. Why they didn't write as $(f(a^3) + f(b^3) + f(c^3) - 3f(a + b + c)) \cdot (a + b + c) \ge 0$ that would've been so sick. Denote both assertions $P(a, b, c)$. Shift such that $f(0) = 0$ and $f(1) = 1$. Note that if $a + b + c = 0$ then it follows that $f(a^3) + f(b^3) + f(c^3) = 3f(abc)$. Claim: $f$ is odd. Proof. By taking $P(a, -a, 0)$ it follows that $f(a) = -f(a)$, so $f(0) = 0$. $\blacksquare$ Claim: $f$ is monotonic. Proof. By plugging in $c = 0$, we get that if $a + b \ge 0$, then $f(a^3) + f(b^3) \ge 0$ and vice versa. It first follows that the sign of $f(a^3)$ is the same as the sign of $a$. It thus follows that $|a| \ge |b| \iff |f(a^3)| \ge |f(b^3)|$, so $f$ is monotonic. $\blacksquare$ Claim: $f$ is continuous. Proof. Now suppose that $r$ is some positive real. FTSOC suppose that $f$ is not continuous at $r$, so there's a jump at $r$. By virtue of being monotonic, $f$ must be continuous at all points but some set of measure $0$. Take $a + b + c = 0$ such $abc = \pm r$, and $f$ is continuous at $a$ and $b$. This is possible by IVT and the density of points with derivatives. Then consider $P(a + \varepsilon, b - \varepsilon, c)$, which preserves the equality. The LHS changes by some arbitrarily small amount, which in turn is equivalent to the RHS changing by some small amount. This gives continuity. $\blacksquare$ Note that we can rewrite the $a + b + c = 0$ condition as \[ f(a^3) + 3f(ab(a+b)) + f(b^3) = f((a+b)^3). \] Claim: $f(2x) = 2f(x)$. Proof. This rearranges as \[ 2f(a^3) + 3f(2a^3) = f(8a^3) \]It then follows by considering the linear recurrence that \[ f(2^n a^3) = c_1 2^n + c_2 (-1)^n + c_3 n (-1)^n \]Since as $n \to -\infty$ on integers, this expression approaches $0$, it must then follow that $c_2 = c_3 = 0$. The result follows. $\blacksquare$ Claim: $f(x) = 3x$ holds as well. Proof. We similarly have that \[ 9f(a^3) + 6f(3a^3) = f(3^3a^3) \]It then follows by the recureence relation \[ f(3^n a^3) = c_1 3^n + c_2 \alpha^n + c_2 ol{\alpha}^n = c_1 3^n + O(3^{n/2}) \]where nonreal $\alpha, \overline{\alpha}$ have magnitude $\sqrt{3}$. Note that $f(2^k) = 2^k$. We first claim that $c_1 = a^3$. Then if $2^k > 3^n a^3$, it follows that \[ 2^k > c_1 3^n + O(3^{n/2}) \]which rearranges as \[ \frac{2^k}{3^n} + O(3^{-\frac{n}{2}}) > c_1 \]Taking $\frac{2^k}{3^n}$ approaching $a^3$ for sufficiently large $n$, it follows that $a^3 \ge c_1$. Likewise, by bounding from below we get that $c_1 \le a^3$, so $c_1 = a^3$. Now, if $2^k > 3^n a^3$ we get that \[ 2^k > 3^n a^3 + c_2 \alpha^n + c_2 \overline{\alpha}^n \]Take extremely negative $k$, $n$ such that $2^k > 3^n a^3$ with error of magnitude $O(3^{n})$. Choose $n \pmod{6}$ to get an error of magnitude $c_2 3^{n/2}$ in the wrong sign, which is a contradiction unless $c_2 = 0$. It thus follows that $f(3x) = 3f(x)$. $\blacksquare$ Since $\log(2), \log(3)$ are independent $\pmod{1}$ by Baker's theorem, it follows then that $f(x) = x$ for all $x$ by approximating $x$ with $2^a3^b$. As such, $f(x) = cx$ for nonnegative $c$ is the solution set.