Trigonometric Inequality with Integrals

Problem

Trigonometric Inequality with Integrals

Solution

Note that

$\begin{align} &|\sin (x-y)\cos (x+y)-\sin (x+y)|\\ &\qquad\qquad\le |\sin (x-y)\cos (x+y)|+|\sin (x+y)|\\ &\qquad\qquad\le |\cos (x+y)|+|\sin (x+y)|\\ &\qquad\qquad\le\sqrt{2}(\cos^2(x+y)+\sin^2(x+y))\\ &\qquad\qquad=\sqrt{2}. \end{align}$

Thus,

$\displaystyle \begin{align}\sum_{cycl}\Omega(a,b)&\le \sqrt{2}\sum_{cycl}\int_a^{2a}\int_b^{2b}1\cdot dxdy\\ &=\sqrt{2}\sum_{cycl}ab\le\sqrt{2}(a^2+b^2+c^2). \end{align}$

Acknowledgment

Dan Sitaru has kindly posted a problem by Dan Sitaru at the CutTheKnotMath facebook page, along with a solution by Ravi Prakash. The problem was previously published at the Romanian Mathematical Magazine.

 

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