parmenides51 wrote: Find all functions $f : R \to R$ such that $f(x^2)-f(y^2) \le (f(x)+y) (x-f(y))$ for all $x, y \in R$.

So basically we have to solve $f:R\to R$ such that $f(x^2)=xf(x)$ and $xy\geq f(x)f(y)$ Following @below completes the proof

$f(x^2)=xf(x)=-xf(-x) \to f(-x)=-f(x)$ $xy \geq f(x)f(y) \to -xy \geq f(-x)f(y)=-f(x)f(y) \geq -xy \to f(x)f(y)=xy $ $f(1)^2=1$ and so $f(x)=\frac{x}{f(1)}=\pm x$

We plug in $x=y=0$, this gives us $x^2 \geq f(x)^2$, thus this implies $x \geq f(x) \geq -x$ Now we set $y=0$ this gives us that $f(x^2) \leq xf(x)$, and setting $x=0$ gives us that $f(y^2) \geq yf(y)$, thus giving us $f(x^2)=xf(x)$ Plugging all of this we get that $f(x)f(y) \leq xy$, but by the first inequality we have that $f(x)f(y) \geq (-x)(-y)=xy$, thus also giving us $f(x)f(y)=xy$. Set $y=1$ we have that $f(x)=\frac{x}{c}$, plugging in the result into $f(x)f(y)$ we get that $c^2 = 1$, thus we get two possible solutions $f(x)=x$ and $f(x)=-x$. Plugging them in into the inequalities we get that both are solutions.

Problem it is from wer

Answer: $f(x)=x$ and $f(x)=-x$ for all real $x$. Proof: $\bullet$ Set $x=y=0:$ $0\leqslant -f(0)^2 \implies f(0)=0.$ $\bullet$ Set $x=0:$ $-f(y^2)\leqslant yf(y)\implies yf(y)\leqslant f(y^2) \ \ \forall y\in \mathbb{R}.$ $\bullet$ Set $y=0:$ $f(x^2)\leqslant xf(x)\ \ \forall y\in \mathbb{R} \implies f(x^2)=xf(x) \ \ \forall x\in \mathbb{R}.$ $\bullet$ Using $f(x^2)=xf(x),$ the original assertion becomes $f(x)f(y)\leqslant xy \ \ \forall x,y\in \mathbb{R}.$ $\bullet$ From $f(x^2)=xf(x)=-xf(-x)$, we find that $f$ is odd. $\bullet$ Set $y=1:$ $f(x)f(1)\leqslant x.$ $\bullet$ Set $y=-1:$ $f(x)f(-1)\leqslant -x \implies f(x)(1)\geqslant x \implies f(x)=\frac{x}{f(1)}.$ $\bullet$ Set $x=y=1:$ $f(1)^2\leqslant 1.$ $\bullet$ Set $x=1, y=-1:$ $-f(1)^2=f(1)f(-1)\leqslant -1 \implies f(1)^2\geqslant 1 \implies f(1)=1 \ \textrm{or} \ -1.$ Thus, $f(x)\equiv x$ and $f(x)\equiv -x$ satisfy our FE.

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.

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

parmenides51 wrote: Find all functions $f : R \to R$ such that $f(x^2)-f(y^2) \le (f(x)+y) (x-f(y))$ for all $x, y \in R$. Nice Problem. Let $P(x, y)$ be the assertion. Then $P(x, x)\implies -x\leq f(x)\leq x \implies f(0)=0$ Now $P(0, y)\implies -f(y^2)\leq -yf(y)$ $P(x, 0)\implies f(x^2)\leq xf(x)$ Hence $f(x^2)=xf(x)$ Now $P(f(y), y)\implies f(f(y)^2)\leq f(y^2)$ $P(x, -f(x))\implies f(x^2)\leq f(f(x)^2)$ Hence $f(f(y)^2)=f(y^2)\implies f(y)f(f(y))=yf(y)\implies f(f(y))=y\implies -f(y)\leq y\leq f(y)$ But $-y\leq f(y)\leq y$ So only $2$ Solutions are there $f(x)\pm x$ $\blacksquare$

Let $P(x,y)$ denote the given assertion. $P(0,0): 0\le -f(0)^2\implies f(0)^2\le 0\implies f(0)=0$. $P(x,0): f(x^2)\le xf(x)$. $P(0,x): -f(x^2)\le -xf(x)\implies f(x^2)\ge xf(x)$. Thus, $f(x^2)=xf(x)$. So $f(x^2)=-xf(-x)\implies xf(x)=-xf(-x)$, since $f(0)=0$, $f$ is odd. The RHS of the original inequality can be written as $xf(x)-f(x)f(y)+xy-yf(y)$. So $xf(x)-yf(y)\le xf(x)-yf(y)+xy-f(x)f(y)\implies xy\ge f(x)f(y)$. Setting $x\to -x$ here gives $-xy\ge -f(x)f(y)\implies xy\le f(x)f(y)$, so $f(x)f(y)=xy$. Setting $y=1$ here gives $f(x)f(1)=x\implies f(x)=\frac{x}{f(1)}$. However, $f(1)^2=1\implies f(1)\in \{-1,1\}$. So $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$, which work.

Let $P(x,y):=f(x^2)-f(y^2)\le(f(x)+y)(x-f(y))$ $P(0,0)$ yields $f(0)^2\le0$ however since a square of any real number is greater than $0$ we must have $0\le f(0)^2\le 0\Longrightarrow f(0)=0$ $P(x,0)$ yields $f(x^2)\le xf(x)$ $P(0,x)$ yields $-f(x^2)\le-xf(x)\Longrightarrow f(x^2)\ge xf(x)$ thus using $P(x,0)$ yields $f(x^2)=xf(x):=Q(x)$ Furthermore $Q(-x)$ yields $f(x^2)=-xf(-x)$ thus $xf(x)=-xf(-x)\Longrightarrow -f(x)=f(-x), \forall x\neq0$ however this is also true for $x=0$ since $f(0)=0$ therefore $f$ is odd. Now notice that expanding the $\text{RHS}$ in $P(x,y)$ yields $f(x^2)-f(y^2)=xf(x)-yf(y)\le xf(x)-f(x)f(y)+xy-yf(y)\Longrightarrow f(x)f(y)\le xy:=T(x,y)$ $T(-x,y)$ yields $f(-x)f(y)\le -xy\Longrightarrow -f(x)f(y)\le -xy\Longleftrightarrow f(x)f(y)\ge xy$ thus $xy\le f(x)f(y)\le xy$ which forces $f(x)f(y)=xy$ Letting $y\to 1$ in the last result we obtain $f(x)f(1)=x\Longrightarrow f(x)=\frac{x}{c}, \forall x\in\mathbb{R}$ where $c=f(1)$ $P(1,1)$ yields $1\ge f(1)^2$ Furthermore $P(1,-1)$ yields $0\le f(1)^2-1\Longrightarrow f(1)^2\ge1$ thus $f(1)^2=1$ which forces $f(1)=\pm 1$ Thus $f(x)=x\text{ or }f(x)=-x$ both of which are solutions to the functional equation. In conclusion $\boxed{f(x)=x\text{ and }f(x)=-x, \forall x\in\mathbb{R}}$ $\blacksquare$.