If not, $f(x+y)\geq{f(xy)}$ holds for every real numbers $x,y$ $P(x,0)$ : $f(x)\geq{f(0)}=c$ where $c$ is a constant. $P(x,-x)$ : $f(0)=c\geq{f(-x^2)}$ , it means that $f(x)=f(0)=c$ for all $x\leq0$ $P(-x,-x)$ where $x\geq0$ : $f(-2x)=c\geq{f(x^2)}$ ,so $f(x)=f(0)=c$ for all $x\geq0$ In conclusion, $f(x)$ is a constant function. Contradiction! So there exists some real numbers $x,y$ satisfying $f(x+y)<f(y)$

What is mean $P(x,0)$

$P(x,y)$ is the statement $f(x+y) \ge f(xy)$. Thus, $P(x,0)$ means $f(x+0)\ge f(x\cdot 0)\implies f(x) \ge f(0)$.

Suppose otherwise, that $f(x+y) \ge f(xy)$ for all $x, y \in \mathbb{R}$. Lemma. For all $a, b \in \mathbb{R}$, if $\frac{a^2}{4} \ge b$, then $f(a) \ge f(b)$. Proof. Since $\frac{a^2}{4} \ge b$, we may take $x=\frac{a+\sqrt{a^2-4b}}{2}$ and $y=\frac{a-\sqrt{a^2-4b}}{2}$ to get $f(a) \ge f(b)$. $\blacksquare$ Since $f$ is non-constant, there are real numbers $s \neq t$ such that $f(s) > f(t)$. But applying the Lemma repeatedly, we have \[f(t) \ge f(-420) \ge f\left(\frac{420^2}{4}\right) \ge f\left(\frac{(420^2/4)^2}{4}\right) \ge \ldots\]where each argument of $f$ (after the second) is $\tfrac14$ of the square of the last. It is clear that these arguments increase without bound, so eventually we will find a term $f(z)$ for which $\frac{z^2}{4} \ge s$. Then \[f(t) \ge \ldots \ge f(z) \ge f(s)\]a contradiction. $\square$

Supose that the statement is not true then take x and y=1 then for all real x $f(x+1)>f(x)$ now take x=1,2,3 to see that $f(6)>f(5)>f(4)>f(3)>f(2)>f(1)$ but we see that f(3x2)>f(2+3) contradiction.

Hi, If the statement is not true, then for all $x$ reals, we yields $f(x+1)\geq f(x)$. Not $f(x + 1) > f (x)$. Unfortunately, we cannot get a contradiction. Also, nothing related to constant function information is used in your solution. Echoes12 wrote: Supose that the statement is not true then take x and y=1 then for all real x $f(x+1)>f(x)$ now take x=1,2,3 to see that $f(6)>f(5)>f(4)>f(3)>f(2)>f(1)$ but we see that f(3x2)>f(2+3) contradiction.

Here is a direct proof: For any nonconstant function $f: \mathbb{R} \to \mathbb{R}$, either i.) $f(0)$ is the absolute minimum or ii.) $f(0)$ is not the absolute minimum. If i.) is true, there there exists $t$ such that $f(0) < f(-t^2)$ or $f(t - t) < f(t \cdot -t)$. Thus, $(x, y) = (t, -t)$ exists in this case. Otherwise if ii.) is true, there exists $t$ such that $f(t) < f(0)$ or $f(t+ 0) < f(t \cdot 0)$. Thus, $(x, y) = (t, 0)$ exists in this case and we are done.

MathRuger wrote: Here is a direct proof: For any nonconstant function $f: \mathbb{R} \to \mathbb{R}$, either i.) $f(0)$ is the absolute minimum or ii.) $f(0)$ is not the absolute minimum. If i.) is true, there there exists $t$ such that $f(0) < f(-t^2)$ or $f(t - t) < f(t \cdot -t)$. Thus, $(x, y) = (t, -t)$ exists in this case. Otherwise if ii.) is true, there exists $t$ such that $f(t) < f(0)$ or $f(t+ 0) < f(t \cdot 0)$. Thus, $(x, y) = (t, 0)$ exists in this case and we are done. Take $f(x) = 0$ where $x \le 0$, and $f(x) = x$ for all $x > 0$, and your argument doesn't work.

Assume FTSOC that $P(x,y):f(x+y)\ge f(xy)\forall x,y\in\mathbb R$. $P(x,0)\Rightarrow f(x)\ge f(0)$ $P(x,-x)\Rightarrow f(-x^2)\le f(0)$ So $f(x)=f(0)\forall x\in\mathbb R^-$. Let $u\in\mathbb R^-$. $P(u,-1)\Rightarrow f(0)\ge f(u)$, so similarly, we have $f$ is constant, contradiction.

Assume not. Then $f(x+y)\ge f(xy)$. Let $P(x,y)$ denote the assertion that $f(x+y)\ge f(xy)$. $P(x,0): f(x)\ge f(0)$. $P(x,-x): f(0)\ge f(-x^2)\ge f(0)$, so $f(-x^2)=f(0)$ for all $x\in \mathbb{R}$, which implies that $f(x)=f(0)$ for all $x\le0$. $P(-x,-1)$ where $x\ge0$ gives $f(-x-1)=f(0)\ge f(x)\ge f(0)$, which implies that $f(x)=f(0)$ when $x\ge0$. Thus, the only solution would be a constant function, which contradicts the problem statement.

MathRuger wrote: Here is a direct proof: For any nonconstant function $f: \mathbb{R} \to \mathbb{R}$, either i.) $f(0)$ is the absolute minimum or ii.) $f(0)$ is not the absolute minimum. If i.) is true, there there exists $t$ such that $f(0) < f(-t^2)$ or $f(t - t) < f(t \cdot -t)$. Thus, $(x, y) = (t, -t)$ exists in this case. Otherwise if ii.) is true, there exists $t$ such that $f(t) < f(0)$ or $f(t+ 0) < f(t \cdot 0)$. Thus, $(x, y) = (t, 0)$ exists in this case and we are done. Follow up on i) If there exist $f(x) = f(0)$ for $x\le0$, then we are done, so assume otherwise. There exist $f(k) > f(0)$ for $k>0$. Take $x=y=-\sqrt{k}$, then we have $f(-2\sqrt{k}) = f(0) < f(k)$ so were done

Assume, for the sake of contradiction, that $f(x + y) \geq f(xy)$ for all $x, y$. $\underline{P}(x, 0) \implies \boxed{f(x) \geq f(0)}$ $\underline{P}(x, -x) \implies f(0) \geq f(-x^2)$. But $f(-x^2) \geq f(0)$! Hence $f(-x^2) = f(0)$, implying that $f(x) = f(0)$ for all negative $x$. $\underline{P}(-x^2, -x^2) \implies f(0) = f(-2x^2) \geq f(x^4)$. But $f(x^4) \geq f(0)$! Hence $f(x^4) = f(0)$, implying that $f(x) = f(0)$ for all positive $x$. Hence $f$ is a constant function, which is a contradiction.