We have $f(x) = x$ and $f(x) = -x,$ so we're done.

Let $P(x,y)$ be the assertion $f(x^2)-f(y^2)\leq (f(x)+y)(x-f(y))$. Then, $P(x,-x)$ gives $$ 0\leq (f(x)-x)(x-f(-x))\Longrightarrow xf(x)-f(x)f(-x)+xf(-x)-x^2\geq 0 $$And $P(-x,x)$ gives $$ 0\leq (f(-x)+x)(-x-f(x)) \Longrightarrow -xf(x)-f(x)f(-x)-xf(-x)-x^2\geq 0 $$Hence, $P(x,-x)+P(-x,x)$ gives $$ -2f(x)f(-x)-2x^2\geq 0 \Longrightarrow 2f(x)f(-x)\leq -2x^2 \qquad(1) $$Also, $P(x,x)$ gives $$ 0\leq (f(x)+x)(x-f(x))\Longrightarrow f(x)^2\leq x^2 \qquad(2) $$And $P(-x,-x)$ gives $$ 0\leq (f(-x)-x)(-x-f(-x))\Longrightarrow f(-x)^2\leq x^2 \qquad (3) $$Adding up $(1), (2)$ and $(3)$ yields $$ f(x)^2+2f(x)f(-x)+f(-x)^2\leq 0 \Longrightarrow [f(x)+f(-x)]^2\leq 0 \Longrightarrow f(x)=-f(-x) $$In other words, $f$ must be odd. But then $(1)$ can be rewritten as $$ -2f(x)^2\leq -2x^2 \Longrightarrow f(x)^2\geq x^2 \qquad(4) $$Now, in light of $(2)$ and $(4)$, it holds $f(x)^2=x^2 \Longrightarrow f(x)=x$ or $f(x)=-x$ for every real $x$ In particular, $f(0)=0$. Now, let's prove that it cannot happen $f(x)=x$ for some non-zero real numbers and $f(x)=-x$ for some other non-zero real numbers. First of all, since $f$ is odd, it holds $$ \left\{\begin{array}{lll} f(x)=x \iff f(-x)=-x \\ f(x)=-x \iff f(-(-x))=-(-x) \\ \end{array} \right. \qquad (5) $$Let's suppose there existed two different non-zero reals $a,b$ such that $f(a)=a$ and $f(b)=-b$. In light of $(5)$, we can suppose $a>0,b<0$. First, computing $P(x,y)+P(y,x)$ gives $$ 0\leq (f(x)+y)(x-f(y))+(f(y)+x)(y-f(x)) \Longrightarrow 0\leq 2xy-2f(x)f(y) \Longrightarrow f(x)f(y)\leq xy \qquad (6) $$In light of $(6)$, $P(a,b)+P(b,a)$ gives $$ f(a)f(b)\leq ab \Longrightarrow -ab\leq ab \Longrightarrow ab\geq 0 $$However, since we were supposing $a>0,b<0$, we have reached a contradiction. Lastly, the two only functions that might yet satisfy the inequality are $f(x)=x$ and $f(x)=-x$. It is easy to check that they follow the original equation, so they are the only two solutions.

@centslordm do u hace proof about these are the only functions that works!!!! Idk if this works, alr lets do this, solved with no one becuase i dont have friends :c Let $P(x,y)$ the assertion of the given F.E. $P(0,0)$ $$-f(0)^2 \ge 0 \implies f(0)=0$$$P(x,0)$ $$xf(x) \ge f(x^2)$$$P(0,x)$ $$xf(x) \le f(x^2)$$By the last ecuations we have that $xf(x)=f(x^2)$ thus its known that $f$ is odd. $P(x,-x)+P(-x,x)$ $$-2x^2 \ge 2f(x)f(-x)=-2f(x)^2 \ge -2x^2 \implies f(x)^2=x^2 \implies f(x)=\pm x$$$P(x,x)$ $$f(x) \ge x \; \text{or} \; f(x) \le -x$$If $f(x) \ge x$ then $f(x)=x$ and if $f(x) \le -x$ then $f(x)=-x$ (there is no need of pointwise trap with this lol) Thus the solutions are $f(x)=x$ and $f(x)=-x$ and we are done

This is Dutch IMO TST Day 3 P2

Let $P(x,y)$ be the given assertion. $P(0,0)\Rightarrow f(0)^2\le0\Rightarrow f(0)=0$ $P(x,0)\Rightarrow f(x^2)\le xf(x)$ $P(0,x)\Rightarrow-f(x^2)\le-xf(x)\Rightarrow f(x^2)\ge xf(x)$ So $f(x^2)=xf(x)$, hence $f$ is odd. Now $P(x,y)$ gives the new assertion $Q(x,y):f(x)f(y)\le xy$. $Q(x,-y)\Rightarrow-f(x)f(y)\le-xy\Rightarrow f(x)f(y)\ge xy$ So equality holds in $Q(x,y)$, so $f(x)=\frac1{f(1)}x$ after setting $y=1$. Then $x=1$ yields the solutions $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$, which both work.

$P(0,0)\Rightarrow f(0)^2\le0\Rightarrow f(0)=0$ $$f(x^2)=xf(x)$$$f$: is odd function $$f(x)f(y)= xy$$$$f(x)=x$$and $$f(x)=-x$$

Asnwer: $f(x)=\pm x \forall x \in \mathbb{R}$ Solution: Let $P(x,y)$-denote the given assertion. We begin by making the following claim: Claim: $f(0)=0$ Proof: $P(0,0,) \implies 0 \leq f(0)^2$ however it's well known that $f(0)^2 \geq 0 \therefore f(0)=0 \square$. Claim: $f(x^2)=xf(x)$ Proof: $P(x,0) \implies f(x^2) \leq xf(x)$ $...(1)$ $P(0,x) \implies -f(x^2) \leq -xf(x) \implies f(x^2) \geq xf(x)$ $...(2)$ Combining $(1)$ and $(2)$ we get: $$xf(x) \stackrel{(1)}{ \geq} f(x^2) \stackrel{(2)}{\geq} xf(x) \implies \boxed{f(x^2)=xf(x)} ...(*) \square$$ Claim: $f-\text{odd}$ Proof: From $(*) \implies f(x^2)=xf(x) , x \rightarrow -x \implies xf(x) \stackrel{(*)}{=} f(x)^2=-xf(x) \implies f(x)=-f(-x) \forall x \in \mathbb{R} / \{0\}.$ Combining with $f(0)=0 \implies f(x)=-f(x) \forall x \in \mathbb{R} \square$. Now by using $(*)$ on $P(x,y)$ we get: $0 \leq - f(x)f(y) +xy \implies f(x)f(y) \leq 0$ denote this by $Q(x,y)$. Claim: $f(x)=\frac{x}{d}$ , $d \neq 0$-constant Proof: Let $f(1)=d$ , $d$-constant. FTSOC assume $d=0$ Let $z \in \mathbb{R^-}$ so $P(z,1) \implies f(z)d \leq z \implies 0 < 0 \rightarrow \leftarrow (\because z \in \mathbb{R^-})$. So $d \neq 0 $ $Q(x,1) \implies f(x)d \leq x...(3)$ Note that since $f-\text{odd}$ we have: $f(1)=-f(-1) \implies f(-1)=-d$ $Q(-1,x) \implies f(x)f(-1) \leq -x \implies -f(x)d \leq -x \implies f(x)d \geq x ...(4)$ Combining $(3)$ and $(4)$ we get: $$x \stackrel{(3)}{\geq} f(x)d \stackrel{(4)}{\geq} x \implies f(x)d=x \implies f(x)=\frac{x}{d} \forall x \in \mathbb{R} \square$$ Claim: $d=1 \vee -1$ Proof: Let $(a,b) \in \mathbb{R^+}$ $Q(a,b) \implies \frac{a}{d} \cdot \frac{b}{d} \leq ab \implies \frac{ab-d^2ab}{d^2} \leq 0 \implies \frac{ab(1-d^2)}{d^2} \leq 0 $ Since $ab \geq 0 , d^2 \geq 0$ we get that $1-d^2 \leq 0 \implies d^2 \geq 1 ...(5)$ $Q(1,1) \implies d^2 \leq 1 ...(6)$ Finally combing $(5)$ and $(6)\implies$ $$1 \stackrel{(5)}{\geq} d^2 \stackrel{(6)}{\geq} 1 \implies d^2=1 \implies d=\pm 1 \square$$ So $f(x)=\pm x$ by $Q(x,x) \implies f(x)^2 \leq x^2$ so if $f(x) \geq x$ we have that $f(x)=x$ otherwise if $f(x) \leq x \implies f(x)=-x$ $\blacksquare$