Maximum of $f(x)=\frac{2x\sqrt{(x+1)}}{(9x^2+3)^{\frac{1}{4}}}+\frac{(1-2x)\sqrt{2-2x}}{(9(1-2x)^2+3)^{\frac{1}{4}}}$ on the interval $[0,1/2]$

Question

I would like to find the maxima of the following function in one variable : $$f(x)=\frac{2x\sqrt{(x+1)}}{(9x^2+3)^{\frac{1}{4}}}+\frac{(1-2x)\sqrt{2-2x}}{(9(1-2x)^2+3)^{\frac{1}{4}}}$$ on the interval $[0,1/2]$. I have already checked that it must occur at $1/3$, but could someone
give me a proper method of proving it (ideally without using computation engines ?). The problem here is obviously that the derivative of this function is not nice-looking at all and i do not wish to "play" with it. Thanks in advance !

Answer 1

By C-S $$f(x)=\frac{2x\sqrt{x+1}}{\sqrt[4]{\frac{1}{4}(1+3)(9x^2+3)}}+\frac{(1-2x)\sqrt{2-2x}}{\sqrt[4]{\frac{1}{4}(1+3)(9(1-2x)^2+3)}}\leq$$ $$\geq\frac{2x\sqrt{x+1}}{\sqrt[4]{\frac{1}{4}(3x+3)^2}}+\frac{(1-2x)\sqrt{2-2x}}{\sqrt[4]{\frac{1}{4}(3(1-2x)+3)^2}}=\frac{2\sqrt2x}{\sqrt3}+\frac{\sqrt2(1-2x)}{\sqrt3}=\sqrt{\frac{2}{3}}.$$ The equality occurs for $x=\frac{1}{3}$,which says that we got a maximal value.

Answer 2

Here's an easier way: notice that $f(x)$ can be rewritten as $$f(x) = 2g(x) + g(1-2x)$$ where $$g(x) = \frac{x\sqrt{x+1}}{(9x^2+3)^\frac 14}$$

if you differentiate $f(x)$, you get $$f'(x) = 2g'(x) -2g'(1-2x)$$ the maxima occurs when $f'(x) = 0$, or $2g'(x) = 2g'(1-2x) \implies x = 1-2x$. Solving this gives us $\boxed{x = \frac 13}$