No Square Is Negative

Let $y=4a-a^4$ .

Take the first derivative and set it to zero: $\frac{dy}{da}=-4a^3+4=0\\a=1$

A local extrema occurs at $a=1$ . Sub in $a=1$ to check. Checking, we find that that is the minimum point. Thus, the answer is 3.

$\begin{aligned} (a^2-1)^2 + 2(a-1)^2 \geq 0 \\ (a^2-1)^2 + 2(a^2-2a+1) & \geq 0 \\ (a^4-2a^2+1) + (2a^2-4a+2) & \geq 0 \\ a^4-4a+3 & \geq 0 \\ 4a-a^4 & \leq 3 \end{aligned}$ Thus, the maximum value is $\boxed 3$

$\begin{aligned} \frac{a^4+1+1+1}{4}&\stackrel{\text{AM-GM}}\geq \sqrt[4]{a^4\times 1\times 1\times 1}\\ \implies a^4&\geq 4a-3 \end{aligned}$ (Note that we can apply AM-GM here, since $a^4\geq 0$ for all real $a$ )

Hence: $4a-a^4\leq 4a-(4a-3)\implies \boxed{4a-a^4\leq 3}$ Equality occurs when $a=1$