Inequalities with Double And Triple Integrals

Problem

Inequalities with Double And Triple Integrals

Solution 1

For all $z\in [0,1]\,$ and $x,y\in\left[0,\frac{\pi}{2}\right],\,$ Jensen's inequality gives

$\cos (zx+(1-z)y)\ge z\cos x+(1-z)\cos y.$

We have

$\displaystyle\begin{align} &\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\cos (zx+(1-z)y)dxdy\\ &\qquad\qquad\ge z\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \cos xdxdy + (1-z)\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\cos y dxdy\\ &\qquad\qquad=z\frac{\pi}{2}+(1-z)\frac{\pi}{2}=\frac{\pi}{2}. \end{align}$

Taking $\displaystyle z=\frac{1}{2}\,$ solves (A).

Further, $\displaystyle \int_0^1\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\cos (xz+y(1-z))dxdy\ge\int_0^1\frac{\pi}{2}=\frac{\pi}{2}\,$ which solves (B).

Solution 2

$\displaystyle \begin{align} &\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\cos\left(\frac{x+y}{2}\right)dxdy =\int_0^{\frac{\pi}{2}}2\left[ \sin\left(\frac{x+y}{2}\right)\right]_0^{\frac{\pi}{2}}dy\\ &\qquad\qquad\qquad=2\int_0^{\frac{\pi}{2}} \left[ \sin \left( \frac{\displaystyle \frac{\pi}{2}+y}{\displaystyle 2}\right)-\sin\left(\frac{y}{2}\right)\right]dy\\ &\qquad\qquad\qquad=2\left[-2\cos \sin \left( \frac{\displaystyle \frac{\pi}{2}+y}{\displaystyle 2}\right)+2\cos\left(\frac{y}{2}\right)\right]_0^{\frac{\pi}{2}}\\ &\qquad\qquad\qquad=4\left[-\sin\frac{\pi}{2}+\sin\frac{\pi}{4}+\cos\frac{\pi}{4}-\cos 0\right]\\ &\qquad\qquad\qquad=4\left[\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-1\right]=4(\sqrt{2}-1)\gt\frac{\pi}{2}. \end{align}$

This solves (A).

Solution 3

$\displaystyle\begin{align} &\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\cos\left(\frac{x+y}{2}\right)dxdy\\ &\qquad\qquad=\int_0^{\frac{\pi}{2}}\cos\frac{x}{2}dx\int_0^{\frac{\pi}{2}}cos\frac{y}{2}dy-\int_0^{\frac{\pi}{2}}\sin\frac{x}{2}dx\int_0^{\frac{\pi}{2}}\sin\frac{y}{2}dy\\ &\qquad\qquad=4\left(\sin^2\frac{\pi}{4}-\left(\cos\frac{\pi}{4}-\cos 0\right)^2\right)\\ &\qquad\qquad=4(\sqrt{2}-1)\gt\frac{\pi}{2}. \end{align}$

This solves (A).

Acknowledgment

Dan Sitaru has kindly posted a problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page. Solution 1 is by Quang Minh Tran; Solution 2 is by Michel Rebeiz; Solution 3 is by Rozeta Atanasova; Diego Alvariz and N. N. Taleb have indepdently come with Solution 1.

 

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