<s>[INST] If $f(x)f(y)+f(xy)\le -\frac{1}{4},\forall x,y\in[0,1)$, show that $f(x)=-\frac{1}{2}$ Let $f:[0,1) \to \mathbb{R}$ be a function such that 
$$f(x)f(y)+f(xy)\le -\dfrac{1}{4} \quad \forall\, x,y\in[0,1).$$
Show that
$$f(x)=-\dfrac{1}{2} \quad \forall\, x \in[0,1).$$
I have proved that $f(0)=-\dfrac{1}{2}$: if $x=y=0$, we have
$$f^2(0)+f(0)\le-\dfrac{1}{4}\Longrightarrow \left( f(0)+\frac{1}{2} \right)^2\le 0\Longrightarrow f(0)=-\dfrac{1}{2}.$$
But I can't prove $f(x)$ be constant. Thanks.
 [/INST] Plugging in $y=0$ gives $f(x) \ge -\frac{1}{2}$ for each $x$. 
Let $y=x$ to get $f(x)^2+f(x^2) \le -\frac{1}{4}$. This implies $f(x^2) \le -\frac{1}{4}$ for each $x$, and so $f(x) \le -\frac{1}{4}$ for each $x$. But then $f(x)^2+f(x^2) \le -\frac{1}{4}$ implies $f(x^2) \le -\frac{1}{4}-(\frac{1}{4})^2 = -\frac{5}{16}$, and so $f(x) \le -\frac{5}{16}$ for each $x$. Doing this again gives $f(x) \le -\frac{1}{4}-(\frac{5}{16})^2 = -\frac{89}{256}$ for each $x$. If we keep doing this, we see that, for any $\epsilon > 0$, $f(x) \le -(\frac{1}{2}-\epsilon)$ for each $x$. Therefore, $f(x) \le -\frac{1}{2}$ for each $x$.  
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