288. An inequality with two A-cevians in a triangle ABC.

by Virgil Nicula, Jun 19, 2011, 1:44 AM

Proposed problem (own). Let $P(x,y,z)$ and $Q(\alpha , \beta ,\gamma )$ be two points with the mentioned barycentrical

coordinates w.r.t. $\triangle ABC$ . Denote $D\in BC\cap AP$ and $S\in BC\cap AQ$ . Prove that the inequality :

$\boxed{2(y+z)(\beta +\gamma )\cdot AD\cdot AS\ \ge\ (y+z)(\beta+\gamma )\left(b^2+c^2-a^2\right)+y\beta \left(a^2+c^2-b^2\right)+z\gamma \left(a^2+b^2-c^2\right)}$ .

Remark. Here are three equivalent forms of upper inequality :

$\blacktriangleright\ 2(\beta +\gamma )\cdot AD\cdot AS\ \ge\ \beta c^2+\gamma b^2+\frac {(\beta +\gamma )\left(zb^2+yc^2\right)-\left(z\beta+y\gamma\right)a^2}{y+z}$ .

$\blacktriangleright\ 2(y+z)(\beta +\gamma )\cdot AD\cdot AS\ \ge\ (y\gamma +z\beta )\left(b^2+c^2-a^2\right)+2\left(z\gamma b^2+y\beta c^2\right)$ .

$\blacktriangleright\ (y+z)(\beta +\gamma )\cdot AD\cdot AS\ \ge\ (y+z)(\beta +\gamma )bc\cdot\cos A+y\beta ac\cdot\cos B+z\gamma ab\cdot\cos C$ .

Examples.

$\blacktriangleright\ Q:=G\ \implies\ 4m_a\cdot AD\ \ge\ b^2+c^2-a^2+\frac {2\left(zb^2+yc^2\right)}{y+z}$ .

$\blacktriangleright\ P:=I$ and $Q:=G\ \implies\ l_am_a\ge\ s(s-a)$ and $l_am_a+l_bm_b+l_cm_c\ge s^2$ .

$\blacktriangleright\ Q:=I\ \implies\ (y+z)(b+c)l_a\cdot AD\ \ge\ 2(yc+zb)s(s-a)$ .

$\blacktriangleright\ P:=L$ (Lemoine) and $Q:=G\ \implies\ s_am_a\ge \frac 14\cdot \left(b^2+c^2-a^2\right)+\frac {b^2c^2}{b^2+c^2}$ , where $s_a=\frac {2bc}{b^2+c^2}\cdot m_a$ .

$\blacktriangleright\ \sum s_am_a\ge \frac 14\cdot\left (a^2+b^2+c^2\right)+\sum\frac {b^2c^2}{b^2+c^2}$ .

$\blacktriangleright$ If $AD$ and $AS$ are isogonal in the vertex $A$ , then $2(y+z)\left(zb^2+yc^2\right)\cdot AD\cdot AS\ \ge\ \left(z^2b^2+y^2c^2\right)\left(b^2+c^2-a^2\right)+4yzb^2c^2$ .
This post has been edited 12 times. Last edited by Virgil Nicula, Nov 21, 2015, 2:58 PM

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