Functional inequality $f(x+y)f(x-y)\leq f^2(x)-f^2(y)$

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be $$f(x+y)f(x-y)\leq f^2(x)-f^2(y)\quad \forall x,y\in\mathbb{R}.$$ Then prove that $$f(x)=-f(-x)$$ and $$f(x+y)f(x-y)=f^2(x)-f^2(y),$$ where $f^2(x)=\big(f(x)\big)^2$.

Source: KMO $1987$

The first is easy. Let the given assertion be $P(x,y)$ $$P(0,0)\implies f^2(0)\leq 0\implies f(0)=0$$ $$P(x,-x)\implies f^2(x)-f^2(-x)\geq 0\tag{*}$$ $$P(0,x)\implies -f^2(x)-f(x)f(-x)\geq 0\tag{**}$$ $$(*)+2(**)\implies -(f(x)+f(-x))^2\geq 0\implies f(x)=-f(-x)$$ But second is not easy. I'm trying to prove $f(x+y)f(x-y)\geq f^2(x)-f^2(y)$. Can anyone help me?

Answer 1

\begin{align*} f(x + y)f(x - y) &= -f(y + x)f(y - x) \\ &\geq -(f^2(y) - f^2(x)) \\ &= f^2(x) - f^2(y) \end{align*}