# Maximise $(x+1)\sqrt{1-x^2}$ without calculus

### 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?

• Or you could go trigonometry way by substituting $x=\cos \theta$ leading to maximise $2\cos ^3(\theta/2)\sin(\theta/2)$, $\theta\in [0,\pi]$ – Rohan Shinde Oct 18 '18 at 4:11

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.
$$\dfrac{1+x+1+x+1+x+3(1-x)}{3+1}\ge\sqrt[4]{(1+x)^33(1-x)}$$
$$(x+1)\sqrt{1-x^2}$$ 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.