# Same Integral, Three Intervals

### Solution 1

Let $\displaystyle \alpha=\arctan \Bigr(\frac{\displaystyle u \sin x}{\displaystyle v+u \cos x}\Bigr)+\arctan \Bigr(\frac{\displaystyle v \sin x}{\displaystyle u+v\cos x}\Bigr).\,$ Then

\displaystyle \begin{align} &\tan \alpha =\frac{\frac{u\sin x}{v+u\cos x}+\frac{b\sin x}{u+b\cos x}}{1-\frac{uv\sin^2 x}{(v+u\cos x)(u+b\cos x)}}\\ &\tan \alpha =\frac{u\sin x(u+v\cos x)+v\sin x(v+u\cos x)}{(v+u\cos x)(u+v \cos x)-uv \sin^2 x}\\ &\tan \alpha =\frac{(u^2+v^2)\sin x+2uv \sin x\cos x}{uv+v^2\cos x+u^2\cos x+uv\cos^2 x-uv\sin^2 x}\\ &\frac{\sin x(u^2+v^2+2uv\cos)}{\cos x(u^2+v^2)+uv(1+\cos 2x)}=\tan \alpha\\ &\frac{\sin x(u^2+v^2+2uv\cos x)}{\cos x(u^2+v^2)+2uv\cos^2 x}=\tan \alpha\\ &\frac{\sin x}{\cos x}=\tan \alpha \Rightarrow \tan \alpha=\tan x\Rightarrow \alpha=x\\ &I(u,v)=\int_u^v x dx=\frac{x^2}{2}\Bigr|_u^v=\frac{v^2-u^2}{2}. \end{align}

Thus we have

$\displaystyle \frac{2}{b-a}I(a,b)+\frac{2}{c-b}I(b,c)+\frac{2}{a-c}I(a,c)=2(a+b+c).$

Suffice it to show that

$\displaystyle 2(a+b+c)\geq \sum \Biggl(\sqrt{ab}+\sqrt{\frac{a^2+b^2}{2}}\Biggl).$

Let $\displaystyle A=\sqrt{\frac{u^2+v^2}{2}};\,$ $\displaystyle B=\sqrt{uv};\,$ $2A^2=u^2+v^2;\,$ $B^2=uv.\,$ Further

\displaystyle \begin{align} &(u+v)^2=u^2+v^2+2uv=2A^2+2B^2\\ &u+v\geq A+B\Leftrightarrow (u+v)^2\geq (A+B)^2\\ &2A^2+2B^2\geq (A+B)^2\Leftrightarrow (A-B)^2\geq 0. \end{align}

It follows that

\displaystyle \begin{align} &a+b\geq \sqrt{ab}+\sqrt{\frac{a^2+b^2}{2}}\\ &b+c\geq \sqrt{bc}+\sqrt{\frac{b^2+c^2}{2}}\\ &c+a\geq \sqrt{ac}+\sqrt{\frac{a^2+c^2}{2}} \end{align}

and, finally,

$\displaystyle 2(a+b+c)\geq \sum \Biggl(\sqrt{ab}+\sqrt{\frac{a^2+b^2}{2}}\Biggl).$

### Solution 2

$\displaystyle I(u,v)=\int_u^v\arctan\left(\frac{u \sin (v)}{u \cos (v)+v}\right)+\arctan\left(\frac{v \sin (v)}{u+v \cos (v)}\right)\, dx.$

We have the following property:

$\displaystyle \arctan(a)+\arctan(b)=\arctan\left(\frac{a+b}{1-a b}\right) + \mathbb{1}_{0 \leq a b \leq 1} \pi$

(note the mistake in Abramowicz & Stigum, p 80)

$\displaystyle \tan\left(\arctan(a)+\arctan(b)\right)= \frac{a+b}{1-a b},$

$\displaystyle a, b \in \left[0,\frac{\pi}{2}\right].$

Allora

$\displaystyle \tan \left(\arctan\left(\frac{u \sin (x)}{u \cos (x)+v}\right)+\arctan\left(\frac{v \sin (x)}{u+v \cos (x)}\right)\right)=\tan (x).$

Since all variables are in $\displaystyle \left(0,\frac{\pi }{2}\right),\,$ $\displaystyle I(u,v)=\frac{v^2}{2}-\frac{u^2}{2},$, the integrand becomes mysteriously $x$, so

$I(a,b)+I(c,a)+I(b,c)= 2(a+b+c).$

We can prove that

$\displaystyle \frac{\sqrt{a^2+b^2}}{\sqrt{2}}+\frac{\sqrt{a^2+c^2}}{\sqrt{2}}+\frac{\sqrt{b^2+c^2}}{\sqrt{2}}+\sqrt{a b}+\sqrt{a c}+\sqrt{b c} -2 (a+b+c)\le 0$

for $\displaystyle a,b,c,\in [0,\frac{\pi}{2}]$, with equality for $a=b=c=1$.

Sidebar

In the process found a potential scary error in the literature. People seem to have suspected it on @StackMath

Riemann Surfaces, sort of. Below is the Abr. & Stig. now used for 50 years!

### Acknowledgment

This is a Dan Sitaru's problem from the Romanian Mathematical Magazine. Dan has kindly sent me the problem and his solution on a LaTeX file, as did N. N. Taleb (Solution 2). I very much apppreciate this kind of thoughtfulness.