48. An inequality in a triangle : h_a+max{b,c} ...

by Virgil Nicula, Jun 28, 2010, 1:02 PM

Let $ ABC$ be a triangle with $a\ge b$ and $a\ge c$ . Prove that $ \boxed {\ h_a + \max\ \{b,c\}\ \le\ 4R\cos^3\frac A3\ }$ where $ h_a$ is the length of the $ A$-altitude

and $ R$ is the circumradius of $ \triangle ABC$ . The trigonometrical form of this inequality is : $ \boxed {\ \cos\ (B-C)+2\max\ \{\sin B,\sin C\}\ \le\ 3\cos\frac A3\ }$ .

Remark. $\ h_a + \max\ \{b,c\}=4R\cos^3\frac A3\ \Longleftrightarrow$ $2B-C=90^{\circ}$ and in this case the rays $[AH$ and $[AO$ trisect $\widehat {BAC}$ , where $H$ - orthocenter and $O$ - circumcenter of $ABC$ .

Particular cases $:\ 1\blacktriangleright\ A = 90^{\circ}\ \implies\ h_a + \max\ \{b,c\}\ \le\ \frac {3a\sqrt 3}{4}\ ;\ 2\blacktriangleright\ A = 135^{\circ}\ \implies\ h_a + \max\ \{b,c\}\ < \ a$ .

Proof (Mateescu Constantin). Suppose w.l.o.g. that $b\ge c$ . Since $0<B-C<\frac {\pi}{2}$ , $h_a=2R\sin B\sin C$ , $b=2R\sin B$ and $\cos A=4\cos^3\frac A3-3\cos\frac A3$ obtain :

$h_a+b\le 4\cos^3\frac A3$ $\Longleftrightarrow$ $2\sin B\sin C+2\sin B\le 4R\cos^3\frac A3$ $\Longleftrightarrow$ $\cos (B-C)+\cos A+2\sin B\le 4\cos^3\frac A3$ $\Longleftrightarrow$ $\cos\ (B-C)+2\sin B\le 3\cos\frac A3$ . Apply the Jensen's

ineq $:\ 2\sin B+\sin\left[\frac{\pi}2-(B-C)\right]\le 3\sin\left(\frac{2B+\frac{\pi}{2}-B+C}{3}\right)\iff$ $2\sin B+\cos (B-C)\ \le\ 3\sin\left(\frac{A+B+C-A+\frac{\pi}{2}}{3}\right)\iff$ $2\sin B+\cos (B-C)\le$

$3\sin\left(\frac{\pi}2-\frac A3\right)=\cos\frac A3.$ We have equality iff $90^{\circ}-(B-C)=B,$ i.e. $2B=90^{\circ}+C.$ In this case $\left\|\begin{array}{c} 2B-C=90^{\circ}\\\\ B+C=180^{\circ}-A\end{array}\right\|$ $\Longleftrightarrow$ $\left\|\begin{array}{c} B=90^{\circ}-\frac A3\\\\ C=90^{\circ}-\frac {2A}{3}\end{array}\right\|$ $\implies$ $B-C=\frac A3.$
This post has been edited 23 times. Last edited by Virgil Nicula, Nov 24, 2016, 3:29 PM

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