$a,b,c \in \mathbb{R}, 0\leq a,b,c \leq 1$, find max $$\frac{a+b+c}{3}+\sqrt{a(1-a)+b(1-b)+c(1-c)}$$
My answer let $t=\dfrac{a+b+c}{3}$, it will be $$t+\sqrt{3t-(x^2+y^2+z^2)}\leq t+\sqrt{3t-3t^2}(\because \mathrm{CS})$$ Then $f(t)=t+\sqrt{3t-3t^2}$, and differentiate to find the extrema
Answer become $a=b=c=\dfrac{3}{4}$, and max is $\dfrac{3}{2}$.
Is there a better way of solving this?
Avoiding derivative you can use CS again:
$$t+\sqrt{3t-3t^2}\leq \sqrt{2(t^2+3t-3t^2)} $$
with eqaulity iff $t^2 = 3t-3t^2$ i.e. $t=0$ or $t={3\over 4}$. Later choise give us $f(t) \leq {3\over 2}$