<s>[INST] Is there an injective function such that $f(x^2)-f^2(x)\ge \frac{1}{4}$? The exercise asks me this:
Is there an injective function such that $f(x^2)-f^2(x)\ge \frac{1}{4}$?
ps: $f: \mathbb{R}\to \mathbb{R}$
I really don't know how to start :c, I appreciate hints.
 [/INST] There is no such injective function. To solve it we first see that if $x^2 = x$ (which is the case for $x=0$ and $x=1$) then $f(x^2) = f(x)$ and $f(x^2) - f^2(x) \geq \frac{1}{4}$ becomes (for $x=0$ or $x=1$)
$$f(x) - f^2(x) - \frac{1}{4} = -\left(f(x)- \frac{1}{2}\right)^2 \geq 0$$
but this is only possible ($-a^2 \geq 0$ implies $a=0$ since a square cannot be negative) if $f(x) = \frac{1}{2}$. We must therefore have $f(0) = f(1) = \frac{1}{2}$ so $f$ cannot be an injection (since for an injective function $f(x) = f(y)$ implies $x=y$).
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