233. An interesting identity and an easy & nice inequality.

by Virgil Nicula, Mar 1, 2011, 7:02 AM

Interesting Identity !

$\boxed{\ \frac{m}{1+m+mn}+\frac{n}{1+n+np}+\frac{p}{1+p+pm}+\frac{(mnp-1)^2}{(1+m+mn)(1+n+np)(1+p+pm)}=1\ }$

Remark. $\boxed{\{m,n,p\}\subset\mathbb R_+^*\ \implies\ \frac{m}{1+m+mn}+\frac{n}{1+n+np}+\frac{p}{1+p+pm}\ \le\ 1}$ with equality iff $mnp=1$ .


A geometrical interpretation. Let $\triangle ABC$ with the area $S=[ABC]=1$ . For the points $\left\{\begin{array}{ccc} M\in (BC) & ; & \frac {MB}{MC}=m\\\\ N\in (CA) & ; & \frac {NC}{NA}=n\\\\ P\in (AB) & ; & \frac {PA}{PB}=p\end{array}\right\|$ define

$\left\{\begin{array}{c} X\in BN\cap CP\\\\ Y\in CP\cap AM\\\\ Z\in AM\cap BN\end{array}\right\|$ . Observe that $[ABZ]+[BCX]+$ $[CAY]+[XYZ]=1$ and prove easily that $[ABZ]=\frac {m}{1+m+mn}$ ,

$[BCX]=\frac {n}{1+n+np}$ , $[CAY]=\frac {p}{1+p+pm}$ and area of $\triangle XYZ$ is $[XYZ]=\frac {(1-mnp)^2}{(1+m+mn)(1+n+np)(1+p+pm)}$ .

Remark. $\sum\frac{m}{1+m+mn}\le 1$ and $\sum \frac{m}{1+m+mn}=1$ $\iff mnp=1\iff AM\cap BN\cap CP\ne \emptyset$ .

Particular case. $m=n=p\implies [XYZ]=\frac {(m-1)^2}{m^2+m+1}\cdot [ABC]$ .


Proposed inequality (easy and nice). $\boxed{\ \{a,b,c\}\subset\mathrm R_+^*\ \implies\ \frac{a^2+bc}{b+c}+\frac{b^2+ca}{c+a}+\frac{c^2+ab}{a+b}\ \ge\ a+b+c\ }$ .

Proof. $\sum{\frac{a^2+bc}{b+c}} \ge a+b+c\iff$ $\sum\left(\frac{a^2+bc}{b+c}+a\right) \ge 2(a+b+c)$ $\iff$ $\sum \frac{(a+b)(a+c)}{b+c}\ge 2(a+b+c)\ (*)$ .

Denote $\left\|\begin{array}{c} x=b+c\\\ y=c+a\\\ z=a+b\end{array}\right\|$ . Thus, the inequality $(*)$ becomes $\sum\frac {yz}{x}\ge x+y+z\iff \sum (yz)^2\ge xyz(x+y+z)$ , what is truly. From Arqady :

$ \sum\left(\frac {a^2+bc}{b+c}-a\right)=$ $\sum\frac {(a-b)(a-c)}{b+c}\ge \frac {(a-b)(a-c)}{b+c}+\frac {(b-c)(b-a)}{c+a}=$ $(a-b)\cdot\left(\frac {a-c}{b+c}-\frac {b-c}{c+a}\right)=$ $\frac {(a-b)^2(a+b)}{(b+c)(c+a)}\ge 0$ .
This post has been edited 54 times. Last edited by Virgil Nicula, Nov 21, 2015, 8:05 AM

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