<s>[INST] Find maximum without calculus Let $f:(0,1]\rightarrow\mathbb{R}$ with $f(x)=2x(1+\sqrt{1-x^2})$. Is it possible to find the maximum of this function without calculus? Possibility through some series of inequalities?
 [/INST] $(f(x))^2 = 4x^2\left(1+\sqrt{1-x^2}\right)^2$. Let $y = \sqrt{1-x^2} \Rightarrow x^2 = 1 - y^2 \Rightarrow (f(x))^2 = 4(1-y^2)(1+y)^2 = 4(1-y)(1+y)^3$. 
Apply AM-GM inequality we have:
$2 = (1-y) + \dfrac{1+y}{3} + \dfrac{1+y}{3} + \dfrac{1+y}{3} \geq 4\sqrt[4]{\dfrac{(1-y)(1+y)^3}{27}} \Rightarrow \dfrac{16\times 27}{4^4} \geq (1-y)(1+y)^3 \Rightarrow \dfrac{27}{16} \geq (1-y)(1+y)^3 \Rightarrow \dfrac{27}{4} \geq 4(1-y)(1+y)^3 = (f(x))^2 \Rightarrow f(x) \leq \sqrt{\dfrac{27}{4}} = \dfrac{3\sqrt{3}}{2} \Rightarrow f_{\text{max}} = \dfrac{3\sqrt{3}}{2}$.
$=$ occurs when $1-y = \dfrac{1+y}{3} \Rightarrow y = \dfrac{1}{2} \Rightarrow \sqrt{1-x^2} = \dfrac{1}{2} \Rightarrow x = \dfrac{\sqrt{3}}{2}$.
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