Prove $a^2\cos B\cos C+b^2\cos C\cos A+c^2\cos A\cos B\leq2S.$

Question

Prove that in any triangle inequality holds:

$$a^2\cos B\cos C+b^2\cos C\cos A+c^2\cos A\cos B\leq2S.$$

Is gender inequality that occurs right triangle, not an equilateral triangle. For this reason I suspect you have used other ways different from the usual.

Answer 1

Let $H$ be the orthocenter of $ABC$. Then $$ d(H,BC) = 2R\cos B\cos C $$ hence: $$ \sum_{cyc} 2R a\cos B\cos C = 2S. $$ If $ABC$ is an acute triangle, then $\cos A,\cos B,\cos C>0$ and $a,b,c < 2R$, hence: $$ \sum_{cyc} a^2\cos B\cos C < \sum_{cyc} 2R a\cos B\cos C = 2S. $$ However, if $ABC$ is an obtuse triangle, the given inequality does not hold: consider, for instance, $(a,b,c)=(\sqrt{3},1,1)$, so that $(A,B,C)=(120^\circ,30^\circ,30^\circ)$, $$\sum_{cyc}a^2\cos B\cos C = \frac{9}{4}-\frac{\sqrt{3}}{2},\qquad 2S=\frac{\sqrt{3}}{2}.$$