Does there exist bounded $f:\mathbb{R} \to \mathbb{R}$ such that $f(1)>0$ and for every $x,y\in \mathbb{R}$ we have $$f(x+y)^2\geq f(x)^2+2f(xy)+f(y)^2$$
I can't solve it. I've put $y=-x$ and similar stuff but doesn't lead anywhere (at least I don't see the way).
Suppose $B<\infty$ is the supremum of $f$. Clearly $B\ge f(1)>0$. Let $g(\epsilon)= f(1)^2+2(B-\epsilon)+(B-\epsilon)^2$. The hypothesis that $f(1)>0$ implies $g(0)>B^2$ and by continuity there exists an $\epsilon>0$ so that $g(\epsilon)>B^2$. For that $\epsilon$ let $x$ be such that $f(x)>B-\epsilon$. Then $$f(1+x)^2\ge f(1)^2+2f(x)+f(x)^2$$ $$ \ge f(1)^2+2(B-\epsilon)+(B-\epsilon)^2=g(\epsilon)>B^2,$$ contradicting the assumption that $B$ was the supremum of $f$.
