274. An inequality in a triangle for its interior point.

by Virgil Nicula, May 8, 2011, 1:34 PM

Proposed problem. Let $ABC$ be a triangle with the circumcircle $w$ . Consider the points

$\left\{\begin{array}{ccc} M\in (BC) & ; & U\in (AM\cap w\\\\ N\in (CA) & ; & V\in (BN\cap w\\\\ P\in (AB) & ; & V\in (CP\cap w\end{array}\right\|$ . Prove that $\boxed{\ \frac {MA}{MU}+\frac {NB}{NV}+\frac {PC}{PW}\ge 9\ }$ .

Proof. Denote $\frac {MB}{MC}=m\ ,\ \frac {NC}{NA}=n\ ,\ \frac {PA}{PB}=p$ . Observe that $\frac {MB}{m}=\frac {MC}{1}=\frac {a}{m+1}$ and $MA\cdot MU=MB\cdot MC$ .

Apply the Stewart's relation for the cevian $AM$ in $\triangle ABC\ :\ a\cdot MA^2+MB\cdot MC\cdot BC=AB^2\cdot MC+AC^2\cdot MB$ $\iff$

$\boxed{(m+1)^2\cdot MA^2=(m+1)\cdot \left(mb^2+c^2\right)-ma^2}\ (*)$ . Therefore, $\frac {MA}{MU}=\frac {MA^2}{MA\cdot MU}=$ $\frac {MA^2}{MB\cdot MC}=$ $\frac {(m+1)^2\cdot MA^2}{(m+1)^2\cdot MB\cdot MC}\stackrel{(*)}{=}$

$\frac {(m+1)\left(mb^2+c^2\right)-ma^2}{ma^2}$ . Therefore, $\left\{\begin{array}{c} \frac {MA}{MU}= (m+1)\cdot\frac {b^2}{a^2}+\frac {m+1}{m}\cdot \frac {c^2}{a^2}-1\\\\ \frac {NB}{NV}=(n+1)\cdot\frac {c^2}{b^2}+\frac {n+1}{n}\cdot \frac {a^2}{b^2}-1\\\\ \frac {PC}{PW}=(p+1)\cdot\frac {a^2}{c^2}+\frac {p+1}{p}\cdot \frac {b^2}{c^2}-1\end{array}\right\|\ \bigoplus\implies$ $\frac {MA}{MU}+\frac {NB}{NV}+\frac {PC}{PW}=$

$\left[(m+1)\cdot\frac {b^2}{a^2}+\frac {n+1}{n}\cdot \frac {a^2}{b^2}\right]+$ $\left[(n+1)\cdot\frac {c^2}{b^2}+\frac {p+1}{p}\cdot \frac {b^2}{c^2}\right]+$ $\left[(p+1)\cdot\frac {a^2}{c^2}+\frac {m+1}{m}\cdot \frac {c^2}{a^2}\right]-3\ge $ $-3+2\cdot\sum \sqrt{\frac {(m+1)(n+1)}{n}}\ge $

$-3+4\cdot\sum\sqrt{\frac {m+1}{n+1}}\ge $ $-3+4\cdot 3=9$ because $\frac {n+1}{n}\ge\frac {4}{n+1}$ a.s.o. and $\sum\sqrt{\frac {m+1}{n+1}}\ge 3\cdot\prod\sqrt{\frac {m+1}{n+1}}=3$ .

Remark. We have equality if and only if $m=n=p=1$ and $a=b=c$ , i.e. the triangle $ABC$ is equilateral and $MNP$ is its median triangle.

If $AM\cap BN\cap CP\ne\emptyset$ , i.e. $mnp=1$ , then prove easily that $\frac {MA}{MU}+\frac {NB}{NV}+\frac {PC}{PW}\ge$ $-3+2\cdot\sum \sqrt{\frac {(m+1)(n+1)}{n}}\ge$

$-3+6\cdot\sqrt[3]{(m+1)(n+1)(p+1)}\ge 9$ because $mnp=1\implies (m+1)(n+1)(p+1)\ge 8$ (using only A.M $\ge$ G.M - inequality).
This post has been edited 27 times. Last edited by Virgil Nicula, Nov 22, 2015, 7:54 AM

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