<s>[INST] 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 $[0,1/2]$ 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 !
 [/INST] 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.
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