First, letting $y=x$ gives $f(0)f(2x)\ge0$, which implies the result unless $f(0)=0$; henceforth suppose so. Next, letting $y=-x$ produces $0\ge f(x)^2-f(-x)^2$ or $f(x)^2\le f(-x)^2$; but swapping $x\mapsto-x$ analogously yields $f(-x)^2\le f(x)^2$, and so $f(x)^2=f(-x)^2~(\ast)$. Then, letting $(x,y)=(x,y),(y,x)$ gives \begin{align*} f(x+y)f(x-y)&\ge f(x)^2-f(y)^2\\ f(y+x)f(y-x)&\ge f(y)^2-f(x)^2, \end{align*}and chaining the $1^\text{st}$ inequality with the negation of the $2^\text{nd}$ produces \begin{align*} f(x+y)f(x-y)\ge-f(x+y)f(y-x)\\ \iff f(w)f(z)\ge-f(w)f(-z),~(\dagger) \end{align*}where in the last step we only redenote $(x+y,x-y)\mapsto(w,z)$ for simplicity. Now, it is given that $(\dagger)$ is strict for some $(\tilde{w},\tilde{z})$: \[f(\tilde{w})f(\tilde{z})>-f(\tilde{w})f(-\tilde{z}).\]But $(\ast)$ implies that the magnitudes of both sides are equal, and so both $f(\tilde{w})f(\tilde{z})$ and $f(\tilde{w})f(-\tilde{z})$ are strictly positive. In particular, $f(\tilde{z})=f(-\tilde{z})\neq0$. Finally, letting $z=\tilde{z}$ in $(\dagger)$ yields $f(w)f(\tilde{z})\ge-f(w)f(-\tilde{z})=-f(w)f(\tilde{z})$, i.e., each $f(w)$ has the same sign as $f(\tilde{z})$. $\square$

Call the assertion $P(x,y).$ If $P(a,b)$ is strict, then $P(a,b)+P(b,a)$ gives $f(a+b)(f(a-b)+f(b-a))>0,$ so $f(a-b)+f(b-a)\ne 0.$ Now $P\left(\frac{x+a-b}2,\frac{x-a+b}2\right)+P\left(\frac{x-a+b}2,\frac{x+a-b}2\right)$ gives $f(x)(f(a-b)+f(b-a))\ge 0.$ If $f(a-b)+f(b-a)>0$ then $f(x)\ge 0$ for all $x,$and if $f(a-b)+f(b-a)<0$ then $f(x)\le 0$ for all $x.$

I am the first proposer of this problem Co-authored with Ivan Chan (who proposed IMO 2023 P3 which imho a lot better than this), and Tristan Chaang (who was the first to supplya clean and neat solution; the original solution by Ivan and me made it looked like it's an A5). I originally thought of the problem with sign flipped, i.e. $f(x + y) f(x - y) \le f(x)^2 - f(y)^2$, and then realized that leads to equality (i.e. no room on what to do). Then I decided to try the other direction and took me a while to notice the fact that became the problem statement! Unfortunately I can't further characterize such functions. (What happens if we have $f(x + y) f(x - y) = f(x)^2 - f(y)^2$ and $f(x_0), f(y_0)$ different signs? How about the case where $f(x + y) f(x - y) > f(x)^2 - f(y)^2$ for all $x, y$? Those characterization seems elusive. )

Here is the corrected solution. Let $P(x,y)$ denote the assertion. $P(x,x)\implies f(2x)f(0) \ge 0$. Now if $f(0) \neq 0 $, then we can just divide both sides by $f(0)$ and switch $x \mapsto \frac{x}{2}$ to get that $f(x) \ge 0$ or $f(x) \le 0$ for all $x$. Otherwise, assume $f(0) = 0$. $P(x,-x) \implies 0 \ge f(x)^2 - f(-x)^2 \implies f(-x)^2 \ge f(x)^2$. Now switching $x \mapsto -x$ in the above inequality, we get that $f(x)^2 \ge f(-x)^2$. Adding these two up, we get $f(-x)^2 + f(x)^2 \ge f(x)^2 + f(-x)^2$ which forces that equality holds in both the cases, i.e., $f(x)^2 = f(-x)^2$ for all $x$. \[ P(x,y) \implies f(x+y)f(x-y) \ge f(x)^2 - f(y)^2. \]\[ P(-y,x) \implies f(-y + x)f(-y-x) \ge f(-y)^2 - f(x)^2 = f(y)^2 - f(x)^2. \]Adding these two up, we get that, \[ f(x-y)(f(x+y) + f(-(x+y))) \ge 0 .\] Now we substitute $x \mapsto \frac{p+q}{2}$ and $y \mapsto \frac{q-p}{2}$ to get, \[ f(p)(f(q) + f(-q)) \ge 0 .\] Now if $f(q)+f(-q) \neq 0$ for some $q$, then we are basically done. FTSOC assume that $f(q) = -f(-q)$ for all $q$, i.e., $f$ is odd. Then, \[ P(x,y) \implies f(x+y)f(x-y) \ge f(x)^2 - f(y)^2 .\] \[ P(y,x) \implies f(y+x)f(y-x) \ge f(y)^2 - f(x)^2 \implies -f(x+y)f(x-y) \ge f(y)^2 - f(x)^2 .\] Adding these two, we get that $0\ge 0$ which forces the equality of both the equations, that is \[ f(x+y)f(x-y) = f(x)^2 - f(y)^2 \]for all $x$, $y$. But this is a contradiction because for $(x_0,y_0)$ the equality fails and we are done. Can someone clarify what the $\gg$ means? @below, thanks

kamatadu wrote: Can someone clarify what the $\gg$ means? I have a solution that seems to be pretty weird . Will be posting soon. Sorry, I meant $\geqslant$, non-strict inequality.

Let $P(x, y)$ denote the given assertion. First, $P(x/2, x/2)$ gives $f(x)f(0) \geq 0$ for all $x$. Thus, we are done if $f(0) \neq 0$. Now, assume $f(0) = 0$. Then, $P(x, -x)$ and $P(-x, x)$ give \[f(-x)^2 \geq f(x)^2 \geq f(-x)^2 \implies f(x)^2 = f(-x)^2 \enspace \forall x \in \mathbb{R}.\]We have \begin{align*} P(\frac{x+y}{2}, \frac{x-y}{2}): f(x)f(y) \geq f\left(\frac{x+y}{2} \right)^2 - f\left(\frac{x-y}{2} \right)^2 \end{align*}for all $x, y \in \mathbb{R}$, and denote this by $Q(x, y)$. Suppose $f(-a) = -f(a)$ for some $a$. Then $Q(x, a)$ and $Q(x, -a)$ yield \begin{align*} f(x)f(-a) =f\left(\frac{x-a}{2} \right)^2 - f\left(\frac{x+a}{2} \right)^2 \enspace \forall x \in \mathbb{R}. \end{align*}Again, replacing $x$ by $-x$ gives $f(-x) f(-a) = -f(x)f(-a)$. It follows that $f(-x) = -f(x)$ for all real $x$, assuming that $f(a) \neq 0$. Then, for any $x, y \in \mathbb{R}$, \begin{align*} f(y+x)f(y-x)\geq f(y)^2 - f(x)^2 \geq -f(x+y)f(x-y)=f(x+y)f(y-x) \end{align*}which forces $P(x, y)$ to be the equality, contradicting the hypothesis. Hence, we now have $f(-x) = f(x)$ for all $x$. It follows that \begin{align*} f(x)f(-y) \geq f\left(\frac{x-y}{2} \right)^2 - f\left(\frac{x+y}{2} \right)^2 \geq -f(x)f(y). \end{align*}That is, $f(x)f(y) \geq 0$ for all real $x, y$. Now, the conclusion follows.

Denote by $P(x,y)$ the assertion of $(x,y)$ into the functional equation. Combining the strict $P(x_0, y_0)$ and $P(y_0, x_0)$, we get: \begin{align*} f(x_0+y_0)f(x_0-y_0)&>f(x_0)^2-f(y_0)^2 \\ f(x_0+y_0)f(y_0-x_0)&\geq f(y_0)^2-f(x_0)^2 \\ \Longrightarrow f(x_0+y_0)\cdot (f(x_0-y_0)+f(y_0-x_0)) &> 0. \end{align*}Note that $f$ works iff $-f$ does, so WLOG $f(s)+f(-s)>0$ where $s = x_0-y_0$. Summing $P(x+s,x)$ and $P(x,x+s)$ now shows $f$ is non-negative, as desired.

Let $(a,b)$ be a pair that makes the inequality strict. Plugging in the pairs $(a,b)$ and $(b,a)$ and summing gives us \[ f(a+b) [ f(a-b) + f(b-a) ] > 0 \implies f(a-b)+f(b-a) \neq 0.\]Let $k$ be any constant; plugging in the pairs $(a+k, b+k)$ and $(b+k, a+k)$ and summing gives us \[ f(a+b+2k) [ f(a-b) + f(b-a) ] \geq 0 \implies \text{sgn} (f(a+b+2k)) = \text{sgn} (f(a-b)+f(b-a))\]Since $k$ varies, $a+b+2k$ can be any real number, so we're done.

We will prove the converse that $f$ achieving both signs implies that the inequality is an equality. Claim: $f(a)$ and $f(-a)$ are of different signs (if they are not both zero) Let $b$ be such that $f(a)$ and $f(b)$ have different signs. Then choose $x$ and $y$ such that $x+y=b$ and $x-y=a$. Then $P(x,y)$ gives that $0>f(x)^2-f(y)^2$ so $P(y,x)$ gives that $f(b)f(-a)$ must be positive, as desired. Claim: $f(0)=0$ If $f(0)\neq 0$ then $P(x,x)$ would contradict our assumption. Claim: $f(a)=-f(-a)$ Compare $P(a,-a)$ and $P(-a,a)$. To finish comparing $P(x,y)$ and $P(y,x)$ shows that the inequality must be an equality.

Let $P(x,y)$ denote the assertion. Rewrite the condition as \[-f(x+y)f(y-x) \le f(x)^2 - f(y)^2 \le f(x+y)f(x-y)\]\[-f(x+y)f(y-x) \le f(x+y)f(x-y)\] Case 1: If there is $r$ such that $f(r) = f(-r) \neq 0$, then we have that by $P(\frac{n+r}{2},\frac{n-r}{2})$, which gives that $-f(n)f(-r) \le f(n)f(r)$ which implies that $f(n)$ has the same sign as $f(r)$ which implies $f(x) \le 0$ or $f(x) \ge 0$. Case 2: If $f(x) = -f(-x)$ for all $x \in \mathbb R$, then we get that $f(x)^2 - f(y)^2 = f(x+y)f(x-y)$ so the inequality is never strict. Having exhausted all cases, we are done.

Let $P(x, y)$ denote the assertion $$f(x+y)f(x-y)\geq f(x)^2-f(y)^2.$$ $P(x_0,y_0)$ and $P(y_0, x_0)$ give $$f(x_0+y_0)f(x_0-y_0)> f(x_0)^2-f(y_0)^2$$and $$f(x_0+y_0)f(y_0-x_0)\geq f(y_0)^2-f(x_0)^2,$$respectively. We add them to get $$f(x_0+y_0)(f(x_0-y_0)+f(y_0-x_0))>0\implies f(x_0-y_0)+f(y_0-x_0)\neq 0.$$ $P\left(\frac{x+x_0-y_0}{2}, \frac{x+y_0-x_0}{2}\right)$ and $P\left(\frac{x+y_0-x_0}{2}, \frac{x+x_0-y_0}{2}\right)$ give $$f(x)f(x_0-y_0)\geq f\left(\frac{x+x_0-y_0}{2}\right)^2-f\left(\frac{x+y_0-x_0}{2}\right)^2$$and $$f(x)f(y_0-x_0)\geq f\left(\frac{x+y_0-x_0}{2}\right)^2-f\left(\frac{x+x_0-y_0}{2}\right)^2,$$respectively. We add them to get $$f(x)(f(x_0-y_0)+f(y_0-x_0))\geq 0,$$so $$f(x_0-y_0)+f(y_0-x_0)>0 \implies f(x)\geq 0$$or $$f(x_0-y_0)+f(y_0-x_0)<0 \implies f(x)\leq 0$$for all $x\in\mathbb{R}$ as desired, and we are done. $\blacksquare$

Let $P(x,y)$ be the assertion of $f(x+y)f(x-y)\ge f(x)^2-f(y)^2$. $P(x_0,y_0) \implies f(x_0+y_0)f(x_0-y_0)>f(x_0)^2-f(y_0)^2$ $P(y_0,x_0) \implies f(x_0+y_0)f(y_0-x_0)\ge f(y_0)^2-f(x_0)^2$ Adding those $2$ inequality yields : $f(x_0+y_0)[f(x_0-y_0)+f(y_0-x_0)]>0$ ...$(1)$ Claim 1. If $f(0)=0$ then $f(x)^2=f(-x)^2$ for all $x\in \mathbb{R}$ Proof. Using $P(x,x)$ we get $f(2x)f(0)\ge 0$. 1) If $f(0)>0$ thus $f(2x)\ge 0 \implies f(x)\ge 0$ for all $x\in \mathbb{R}$. 2) If $f(0)<0 $ thus $f(2x)\le 0 \implies f(x) \le 0$ for all $x \in \mathbb{R}$. 3) If $f(0)=0$ then $P(x,-x) \implies f(x)^2\le f(-x)^2$ $P(-x,x) \implies f(x)^2\ge f(-x)^2$ Therefore, $f(x)^2=f(-x)^2 ,\forall x \in \mathbb{R}$. $\square$ By substituting $x \to x_0-y_0$ we have $f(x_0-y_0)^2=f(y_0-x_0)^2$ Therefore, $f(x_0-y_0)=f(y_0-x_0)$ or $f(x_0-y_0)=-f(y_0-x_0)$. Now, consider inequality $(1)$. If $f(x_0-y_0)=-f(y_0-x_0) \implies 0>0$, contradiction. Therefore, $f(x_0-y_0)=f(y_0-x_0)$. Claim 2. If there exist $a\in \mathbb{R}$ such that $f(a)=f(-a)\neq 0$, then $f(x)\ge 0$ or $f(x)\le 0$ for all $x \in \mathbb{R}$ Proof. $P(\frac{a+b}{2},\frac{a-b}{2}) \implies f(a)f(b)\ge f(\frac{a+b}{2})^2-f(\frac{a-b}{2})^2$ $P(\frac{-a+b}{2},\frac{-a-b}{2}) \implies f(-a)f(b) \ge f(\frac{b-a}{2})^2-f(\frac{-a-b}{2})^2=f(\frac{a-b}{2})^2-f(\frac{a+b}{2})^2$ Adding those $2$ inequality yields : $f(b)[f(a)+f(-a)]\ge 0$ Now, if $f(a)=f(-a)>0$ then $f(b)\ge 0, \forall b \in \mathbb{R} \implies f(x)\ge 0, \forall x \in \mathbb{R} $. If $f(a)=f(-a)<0$ then $f(b)\le 0, \forall b \in \mathbb{R} \implies f(x)\le 0, \forall x \in \mathbb{R}$. $\square$ Now, consider the information $f(x_0-y_0)=f(y_0-x_0)$. We claim that $f(x_0-y_0)=f(y_0-x_0)\neq 0$. Assume $f(x_0-y_0)=f(y_0-x_0)=0$. $P(x_0,y_0) \implies f(x_0)^2 < f(y_0)^2$ $P(y_0,x_0) \implies f(x_0) \ge f(y_0)^2$ Contradiction. Therefore $f(x_0-y_0)=f(y_0-x_0)\neq 0$. Therefore, there exist $a=x_0-y_0$ such that $f(a)=f(-a)\neq 0$. Using Claim $2$, we are finished. Done. $\blacksquare$

Guess who didn't know what strict meant. Note that if $f$ is a solution then $c \cdot f$ is a solution. As such, WLOG scale such that $f(1) = 1$. Denote the assertion with $P(x, y)$. By $P(x, x)$ we get that $f(2x)f(0) \ge 0$ which finishes if $f(0) \ne 0$, so assume $f(0) = 0$. Now, by $P(x, -x)$ we get that \[ 0 \ge f(x)^2 - f(-x)^2 \]Comparing with $P(-x, x)$, we get that $f(x)^2 = f(-x)^2$. Claim: $f$ is either even or odd. Proof. Suppose that $f(a) = f(-a) \ne 0, f(b) = -f(-b) \ne 0$. Then set $x + y = a, x - y = b$. Now $P(x, y), P(-x, -y)$ have LHS with opposite signs so $f(x)^2 - f(y)^2 \le 0$. By comparing $P(y, x), P(-y, -x)$ we get $f(y)^2 - f(x)^2 \le 0$ as well. As such, $f(x)^2 = f(y)^2$ and thus $f(a)f(b) \ge 0, f(-a)f(-b) \ge 0$, contradiction. $\blacksquare$ First suppose $f$ is odd. Note that by $P(y, x)$ we have that $f(x + y)f(y - x) \ge f(y)^2 - f(x)^2$. Since $f(x + y)f(y - x) = -f(x + y)f(y - x) \le -(f(x)^2 - f(y)^2)$ it follows that equality must hold for all $x, y$, contradiction. Thus, $f$ is even. Take $x_0, y_0$ such that $f(x_0 + y_0)f(x_0 - y_0) > f(x_0)^2 - f(y_0)^2$. Now, suppose $f(a)$ has a negative sign. Then take $x + y = a, x - y = 1$. We then get that \[ f(a) \ge f(x)^2 - f(y)^2 \]By $P(y, x)$, we also get \[ f(a) \le f(x)^2 - f(y)^2 \]This gives a contradiction.

Anzoteh wrote: Unfortunately I can't further characterize such functions. (What happens if we have $f(x + y) f(x - y) = f(x)^2 - f(y)^2$ and $f(x_0), f(y_0)$ different signs? How about the case where $f(x + y) f(x - y) > f(x)^2 - f(y)^2$ for all $x, y$? Those characterization seems elusive. ) The problem with the sign being equal was actually already an existing problem from IMOR 2018, proposed by Nuno Arala and Miguel Moreira from Portugal. https://artofproblemsolving.com/community/c6h1673292p10651933 And yes there are unexpected extraneous solutions.

Let $P(x,y)$ denote the given assertion. Suppose that the result was not true and choose reals $a,b$ such that $f(a) < 0$ and $f(b) > 0$. Additionally, choose $c,d$ such that $f(c+d) f(c-d) > f(c)^2 - f(d)^2$. Claim: $f(0) = 0$ Proof: Suppose otherwise. $P(x,x): f(2x) f(0) \ge 0$. However, choosing $x = \frac{a}{2}$ implies that $f(a) f(0) \ge 0$, so $f(0) < 0$, but $x = \frac b2$ implies that $f(b) f(0) \ge 0$, so $f(0) > 0$, absurd. Thus, $f(0)$ must equal $0$. $\square$ $P(x,-x): 0 \ge f(x)^2 - f(-x)^2$. But $P(-x,x)$ gives $0\ge f(-x)^2 - f(x)^2$, so this means $f(x)^2 = f(-x)^2$, so either $f(x) = f(-x)$ or $f(x) = -f(-x)$. Claim: $f(c-d) \ne -f(d-c)$ Proof: Suppose that $f(c-d) = -f(d-c)$. Notice that $f(c+d) f(c-d) > f(c)^2 - f(d)^2$ and $P(d, c)$ gives that $f(c+d)f(d-c) \ge f(d)^2 - f(c)^2$. Now, taking the negative of both sides and flipping the inequality gives that $f(c+d) f(c-d) \le f(c)^2 - f(d)^2$, absurd. $\square$ This implies that $f(c-d) = f(d-c)$. $P \left( \frac{x + c - d}{2}, \frac{x - (c - d) }{2} \right): f(x) f(c-d) \ge f \left( \frac{x + c - d}{2} \right)^2 - f\left( \frac{x - (c-d)}{2} \right)^2 $ $P \left( \frac{x - (c - d)}{2} , \frac{x + c + d}{2} \right): f(x) f(d-c) = f(x) f(c-d) \ge f\left( \frac{x - (c-d)}{2} \right)^2 - f \left( \frac{x + c - d}{2} \right)^2$. This implies if $k = f \left( \frac{x + c - d}{2} \right)^2 - f\left( \frac{x - (c-d)}{2} \right)^2$, then $f(x) f(c-d) \ge k$ and $f(x) f(c-d) \ge -k$, so $f(x) f(c-d) \ge 0$. However, if $f(c-d) = 0$, then $f(d-c) = 0$, contradiction to our claim that $f(c-d) \ne -f(d-c)$. Therefore, $f(c-d) \ne 0$, choosing $x\in \{a,b\}$ such that $f(x)$ and $f(c-d)$ have opposite signs gives that $f(x) f(c-d) < 0$, contradiction. Hence we must have $f(x) \ge 0$ always or $f(x) \le 0$ always.

Interesting FE.

Okay retirement days are kinda interesting. I thought I forgot to flip the sign of the inequality somewhere and fakesolved. Turns out I was right all along! Here's a solution which should be right I think. Solution: Denote $P(x,y)$ as the assertion to the inequality. With the usual swapping trick, consider $P(x,y)$ and $P(y,x)$. This will yield the following two inequalities. \begin{align*} f(x+y)f(x-y) &\ge f(x)^2 - f(y)^2 \\ f(x+y)f(y-x) &\ge f(y)^2 - f(x)^2 \end{align*}Add them up and write $y = x + a$ for some arbitrary real $a$. \[f(2x+a) \left(f(-a) + f(a) \right) \ge 0.\]If $f$ is not an odd function, then we can find some $a_0 \in \mathbb{R}$ such that $f(a_0) + f(-a_0) \ne 0$. Then dividing the above inequality by $\left(f(a_0) + f(-a_0)\right)$ with an appropriate change in sign of inequality, we would be done. Henceforth assume $f$ is an odd function. Since $f$ is odd, $f(x-y) = -f(y-x)$. From the above aligned equations, we can now say that \[f(x+y)f(x-y) = f(x)^2 - f(y)^2 \]for all $x,y \in \mathbb{R}$ which is the desired contradiction since there exists some $x_0,y_0 \in \mathbb{R}$ such that there is strict inequality in $P(x,y)$. This completes the solution. $\blacksquare$

OG, Let $P(x,y)$ denote the given assertion. $P(x,x) \implies$ $f(2x)f(0) \geq 0.$ If $f(0)$ is not 0, then the problem statement is directly implied, Hence, suppose $f(0)=0$ $P(x,-x) \implies$ $(f(x))^2 \geq (f(-x))^2$, replacing $x$ by $-x$ $\implies$ $(f(x))^2 \leq (f(-x))^2$. Thus, $(f(x))^2 =(f(-x))^2$ $\implies$ $f(x)= f(-x) or -f(-x)$. Now, Note that the problem condition can be translated to $f(x)f(y) \geq (f(\frac{x+y}{2}))^2-(f(\frac{x-y}{2}))^2$ for every $x,y\in\mathbb{R}$ Claim: $f(x)=f(-x)$ for all real $x$ Assume a real number $z /neq 0$(as $f(0)=0$) exists such that $f(z)=-f(-z)$. Choose $x$ and $y$ such that and $x-y=z$ $P(x,y) \implies$

Let $P(x, y)$ denote the assertion. \[P(x_0, y_0)+P(y_0, x_0)\]\[f(x_0+y_0)(f(x_0-y_0)+f(y_0-x_0))>0\]Let $s=x_0-y_0$. \[P(x, x+s)+P(x+s, x)\]\[f(2x+s)(f(s)+f(-s))\geq 0\]Thus for all $x$ we have the desired result as $(f(s)+f(-s))>0$.

Clearly $f$ is not identically 0; since $f$ satisfies the FE iff $-f$ satisfies the FE, WLOG let $f(1)>0$, and denote the FE as $P(x, y)$. $P(x/2, x/2)\implies f(x)f(0) \ge 0$, so either $f(x)\ge 0$ in which we're done, or $f(0)=0$. $P(x, -x)\implies 0 \ge f(x)^2-f(-x)^2\implies f(x)^2\le f(-x)^2$. But $f(-x)^2\le f(x)^2$ as well so $|f(x)|=|f(-x)|\quad (*)$. $P(x, y) + P(y, x)\implies f(x+y)(f(x-y) + f(y-x))\ge 0$, and if we choose such that $x+y=1$ and $x-y=z \in \mathbb{R}$, then $f(z)+f(-z)\ge 0$ for all $z\in \mathbb{R}$, which by $(*)$ implies $f(z)\ge 0$ or $-f(z)=f(-z)\ne 0$. Let the set of all $z$ that satisfy the latter be $N$; note that if $|N|=0$ we're done. Note that if $x-y$ satisfies $-f(x-y)=f(y-x)$, then $f(x+y)f(x-y) = -f(x+y)f(y-x)\ge f(x)^2-f(y)^2\implies f(x+y)f(y-x)\le f(y)^2-f(x)^2$. But $P(y, x) \implies f(x+y)f(y-x)\ge f(y)^2-f(x)^2$ already so $f(x+y)f(y-x)=f(y)^2-f(x)^2$. But substituting $(x, y) \to (-x, -y)$, we similarly get $f(-x-y)f(x-y) = f(y)^2-f(x)^2\implies f(-x-y)f(y-x) = f(x)^2-f(y)^2\quad (**)$. If we choose to further specify that $x-y\in N$, adding with the previous equality, gives $$(f(x+y)+f(-x-y))f(y-x) = 0\implies -f(x+y)=f(-(x+y))$$as $f(y-x) \ne 0$. Thus, if $|N|>0$, then $f(x)=-f(x)$ for all $x\in \mathbb{R}$. This finally gives a contradiction as this would imply, by $(**)$, that $f(-x-y)f(y-x)=f(x)^2-f(y)^2$ for all $x, y\in \mathbb{R}$; yet the problem statement says there should exist $x_0, y_0$ for which this is not true. Thus, $|N|=0$ and we conclude $f(z)\ge 0$ for all $z\in \mathbb{R}$. $\blacksquare$

Putting $y=x$ gives us $f(2x)f(0) \geq 0 $. This gives $f(0)=0$, if not, then for all $x$, the sign of $f(x)$ doesn't change. Then setting $x=-y$ gives $0\geq f(-y)^2-f(y)^2$, changing $y$ with $ -y$ gives us $f(y)^2\geq f(-y)^2\geq f(y)^2$, which implies $|f(x)|=|f(-x)|$ and $f(x)^2=f(-x)^2$. Then swapping $-x$ as $x$ gives $f(y-x)f(-x-y)\geq f(x)^2-f(y)^2$ . Swapping $x$ and $y$ gives $f(x-y)f(-x-y)\geq f(y)^2-f(x)^2$ which means $f(x)^2-f(y)^2\geq -f(x-y)f(-x-y)$. Assume $f(a)>0>f(b)$. Choose $x$ and $y$ as $x-y=a$, $x+y=c$, where c is any real number. $f(x+y)f(x-y)\geq f(x)^2-f(y)^2\geq -f(x-y)f(-x-y)$ $\Rightarrow f(c)f(a) \geq -f(a)f(-c)$ $\Rightarrow f(c) \geq -f(-c)$ Putting $-c$ as $c$ gives $f(c) \geq -f(-c) \geq f(c)$ which implies $f(c)=-f(-c)$. It is given that $f(x_0+y_0)f(x_0-y_0)>f(x_0)^2-f(y_0)^2$ , $f(y_0+x_0)f(y_0-x_0)\geq f(y_0)^2-f(x_0)^2$. So, $f(x_0+y_0)f(x_0-y_0)>f(x_0)^2-f(y_0)^2 \geq f(x_0+y_0)f(x_0+y_0)$, which gives us contradiction. $\blacksquare$

kawaii

By setting $x=y$, $f(2x)f(0) \geq 0$. Thus, if $f(0) \neq 0$, we may assume without loss of generality that $f(0) > 0$, from where $f(2x) \geq 0$ for all real numbers $x$, so the desired result is true. If $f(0) = 0$, then \[f(x)^2-f(-x)^2 \leq f(0)f(2x) = 0.\]In particular, swapping $x$ and $-x$, it follows that $f(x)^2 = f(-x)^2$. Now, summing the equations for $(x, y)$ and $(y, x)$, \[f(x+y)(f(x-y) + f(y-x)) \geq f(x)^2-f(y)^2+f(y)^2-f(x)^2 = 0.\]In particular, there is $(x_0, y_0)$ such that this inequality is strict; i.e. for this pair $(x_0, y_0)$, it follows that $f(c) = f(-c) \neq 0$ where $c = x_0-y_0$. However, for all pairs $(x, x+c)$, it follows that \[f(2x+c)(f(c)+f(c))) \geq 0\]by the same argument, hence if $f(c) > 0$, $f(2x+c) \geq 0$ for all $x$ and vice versa.