An Inequality with Sides, Cosines, and Semiperimeter

Solution 1

The problem is the same as

$\displaystyle \sum_{cycl}a^2(b\cos B+c\cos C)\le \frac{(a+b+c)^3}{9}.$

Due to the AM-GM inequality, suffice it to prove that

$\displaystyle \sum_{cycl}a^2(b\cos B+c\cos C)\le 3abc.$

We shall show that in fact

$\displaystyle \sum_{cycl}a^2(b\cos B+c\cos C)= 3abc.$

We'll prove that identity in the form $\displaystyle \sum_{cycl}a^2(b^2+c^2)(2bc\cos A)=6a^2b^2c^2.\,$ Indeed,

\displaystyle \begin{align} &\sum_{cycl}a^2(b^2+c^2)(2bc\cos A)=\sum_{cycl}a^2(b^2+c^2)(b^2+c^2-a^2)\\ &\qquad\qquad=\sum_{cycl}[a^2(b^2+c^2)^2 -\sum_{cycl}a^4(b^2+c^2)]\\ &\qquad\qquad=\sum_{cycl}[a^2b^4+a^2c^4+2a^2b^2c^2] -\sum_{cycl}[a^4b^2+a^4c^2]\\ &\qquad\qquad=6a^2b^2c^2. \end{align}

Solution 2

As in Solution 1, we aim to prove

$\displaystyle \sum_{cycl}a^2(b\cos B+c\cos C)= 3abc.$

This is equivalent to

$\displaystyle 2\sum_{cycl}\sin^2 A(\sin 2B+\sin 2C)=12\sin A\sin B\sin C.$

And further,

\displaystyle\begin{align} &2\sum_{cycl}\sin^2 A(\sin 2B+\sin 2C)\\ &\qquad\qquad=\sum_{cycl}(1-\sin 2A)(\sin 2B + \sin 2C)\\ &\qquad\qquad= 2\sum_{cycl}\sin 2A-\sum_{cycl}\sin (2B+2C)\\ &\qquad\qquad= 2\sum_{cycl}\sin 2A+\sum_{cycl}\sin 2A=3\sum_{cycl}\sin 2A\\ &\qquad\qquad= 12\sin A\sin B\sin C, \end{align}

Acknowledgment

Dan Sitaru has kindly posted the problem at the CutTheKnotMath facebook page. This problem of his was originally published at the Romanian Mathematical Magazine. Solutions 1 and 2 are by Kevin Soto Palacios. Amit Itagi independently came up with Solution 1.