Maximum of three variables function

Question

Let $a,b,c\in[0,1]$ and such that $a\leq b\leq c$. Find the maximum of $$ f(a,b,c)=c(\sqrt{a}+\sqrt{b})-(a+b)\sqrt{c} $$ This is a problem in the contest for talented students. I have tried but it is extremely difficult.

Attempt Observe that $f(a,b,c)\leq \sqrt{a}+\sqrt{b}-a-b$. Indeed, the latter inequality is equivalent to $$ (1-\sqrt{c})((\sqrt{a}+\sqrt{b})(1-\sqrt{c}-a-b))\geq 0. $$

Answer 1

Set $x = \sqrt a, y = \sqrt b, z = \sqrt c$. Then $0 \le x \le y \le z \le 1$, and $$ f(a, b, c) = z^2(x+y) - (x^2 + y^2) z = z x (z-x) + z y (z - y) \, . $$ Applying the inequality of arithmetic and geometric means we get that $$ f(a, b, c) \le z \left( \frac{x + (z-x)}{2}\right)^2 + z \left( \frac{y + (z-y)}{2}\right)^2 = \frac{z^3 }{2} \le \frac 12. $$ Equality holds for $(x, y, z) = (\frac 12, \frac 12, 1)$, that is for $(a, b, c) = (\frac 14, \frac 14, 1)$.

Answer 2

$$\frac{\partial^2 f}{\partial c^2}=\frac{a+b}{4\sqrt{c^3}}>0.$$

Thus, $f$ is a convex function of $c$ and $$\max{f}=\max\{f(a,b,1),f(a,b,b)\}.$$ Now, by AM-GM: $$f(a,b,1)=\sqrt{a}+\sqrt{b}-a-b=\sqrt{a}(1-\sqrt{a})+\sqrt{b}(1-\sqrt{b})\leq$$ $$\leq\left(\frac{\sqrt{a}+1-\sqrt{a}}{2}\right)^2+\left(\frac{\sqrt{b}+1-\sqrt{b}}{2}\right)^2=\frac{1}{2},$$ where the equality occurs for $a=b=\frac{1}{4}.$

Also, $$f(a,b,b)=\sqrt{ba}(\sqrt{b}-\sqrt{a})\leq\sqrt{b}\left(\frac{\sqrt{a}+\sqrt{b}-\sqrt{a}}{2}\right)^2=\frac{1}{4}b\sqrt{b}<\frac{1}{2},$$ which says that $\frac{1}{2}$ is a maximal value.