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Problem

Maximise $f:[-1,1]\rightarrow \mathbb{R}$, with $f(x)=(1+x)\sqrt{1-x^2}$

With calculus, this problem would be easily solved by setting $f'(x)=0$ and obtaining $x=\frac{1}{2}$, then checking that $f''(\frac{1}{2})<0$ to obtain the final answer of $f(\frac{1}{2})=\frac{3\sqrt{3}}{4}$

The motivation behind this function comes from maximising the area of an inscribed triangle in the unit circle, for anyone that is curious.

My Attempt

$$f(x)=(1+x)\sqrt{1-x^2}=\sqrt{(1-x^2)(1+x)^2}=\sqrt 3 \sqrt{(1-x^2)\frac{(1+x)^2}{3}}$$

By the AM-GM Inequality, $\sqrt{ab}\leq \frac{a+b}{2}$, with equality iff $a=b$

This means that

$$\sqrt 3 \sqrt{ab} \leq \frac{\sqrt 3}{2}(a+b)$$

Substituting $a=1-x^2, b=\frac{(1+x)^2}{3}$,

$$f(x)=\sqrt 3 \sqrt{(1-x^2)\frac{(1+x)^2}{3}} \leq \frac{\sqrt 3}{2} \left((1-x^2)+\frac{(1+x)^2}{3}\right)$$

$$=\frac{\sqrt 3}{2} \left(\frac{4}{3} -\frac{2}{3} x^2 + \frac{2}{3} x\right)$$

$$=-\frac{\sqrt 3}{2}\frac{2}{3}(x^2-x-2)$$

$$=-\frac{\sqrt 3}{3}\left(\left(x-\frac{1}{2}\right)^2-\frac{9}{4}\right)$$

$$\leq -\frac{\sqrt 3}{3}\left(-\frac{9}{4}\right)=\frac{3\sqrt 3}{4}$$

Both inequalities have equality when $x=\frac{1}{2}$

Hence, $f(x)$ is maximum at $\frac{3\sqrt 3}{4}$ when $x=\frac{1}{2}$

However, this solution is (rather obviously I think) heavily reverse-engineered, with the two inequalities carefully manipulated to give identical equality conditions of $x=\frac{1}{2}$. Is there some better or more "natural" way to find the minimum point, perhaps with better uses of AM-GM or other inequalities like Jensen's inequality?

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  • $\begingroup$ Or you could go trigonometry way by substituting $x=\cos \theta$ leading to maximise $2\cos ^3(\theta/2)\sin(\theta/2)$, $\theta\in [0,\pi]$ $\endgroup$ – Rohan Shinde Oct 18 '18 at 4:11
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By AM-GM $$(1+x)\sqrt{1-x^2}=\frac{1}{\sqrt3}\sqrt{(1+x)^3(3-3x)}\leq$$ $$\leq\frac{1}{\sqrt3}\sqrt{\left(\frac{3(1+x)+3-3x}{4}\right)^4}=\frac{3\sqrt3}{4}.$$ The equality occurs for $1+x=3-3x,$ which says that we got a maximal value.

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$$\dfrac{1+x+1+x+1+x+3(1-x)}{3+1}\ge\sqrt[4]{(1+x)^33(1-x)}$$

Square both sides

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You can solve the problem using geometry.

$(x+1)\sqrt{1-x^2}$ is the area of triangle $$(-1, 0), (x, \sqrt{1-x^2}), (x, -\sqrt{1-x^2}).$$ This unit inscribed triangle has maximal area if and only if it is a equilateral triangle.

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