12. Inegalitatea I. Maftei & S. Radulescu (2).

by Virgil Nicula, Apr 20, 2010, 1:34 AM

http://www.artofproblemsolving.com/viewtopic.php?t=289839

Lemma 1. In any acute triangle $ ABC$ there is the inequality $ \boxed{\ \frac {a^3+b^3+c^3}{abc}\ \ge\ 1+\frac Rr\ }\ \ (1)$ .

Proof. Apply the Walker's inequality, i.e. $ s^2\ge 2R^2+8Rr+3r^2\ (*)$ . Indeed, $ \frac {a^3+b^3+c^3}{abc}=3+\frac {a^3+b^3+c^3-3abc}{abc}=$

$ 3+\frac {(a+b+c)\left[\left(a^2+b^2+c^2\right)-(ab+bc+ca)\right]}{abc}=$ $ 3+\frac {s^2-3r^2-12Rr}{2Rr}=$ $ \frac {s^2-3r^2-6Rr}{2Rr}\ \stackrel{(*)}{\ge}\ \frac {\left(2R^2+8Rr+3r^2\right)-3r^2-6Rr}{2Rr}=$ $ 1+\frac Rr$ .

Lemma 2. In any acute triangle $ ABC$ there is the inequality $ \boxed {\ \sin\frac A2+\sin\frac B2+\sin\frac C2\ \ge\ \frac 54+\frac {r}{2R}\ }\ \ (2)$ .

Proof. Apply the Popoviciu's inequality to the concave function $ \mathrm{COS}$ on the interval $0<x<\frac {\pi}2\ :\ \cos A+\cos B+\cos C+$ $ 3\cos\frac{A+B+C}{3}\ \le$

$2\cdot\left(\cos\frac{A+B}{2}+\cos\frac{B+C}{2}+\cos\frac{C+A}{2}\right)\ \Longleftrightarrow$ $ 1+\frac rR+$ $ \frac 32\ \le\ 2\cdot\left(\sin\frac A2+\sin\frac B2+\sin\frac C2\right)\ \Longleftrightarrow\ \sin\frac A2\ +\ \sin\frac B2\ +\ \sin\frac C2\ \ge\ \frac 54\ +\ \frac r{2R}$ .

PP1. Prove that in any acute triangle $ ABC$ there is the inequality $ \boxed {\ \sum\sqrt {\frac {b+c-a}{a}}\ \ge\ \sqrt {5+\frac Rr+\frac {4r}{R}}\ \ge 3\ }$ .

Proof. $ \left(\sum\sqrt {\frac {b+c-a}{a}}\right)^2=$ $ \sum\frac {b+c-a}{a}+2\cdot\sum\sqrt{\frac {(a+b-c)(a+c-b)}{bc}}=$ $ \sum\left(\frac ab+\frac ba\right)-3+4\cdot\sum\sqrt {\frac {(s-b)(s-c)}{bc}}=$

$ \sum\frac {c^2+2ab\cdot\cos C}{ab}-3+4\cdot\sum\sin\frac A2=$ $ \sum\frac {c^2}{ab}+2\sum\cos C-3+4\cdot\sum\sin \frac A2=$ $ \frac {a^3+b^3+c^3}{abc}+$ $2\cdot\left(1+\frac rR\right)-3+$ $4\cdot\sum\sin\frac A2 \stackrel{(1 \wedge 2)}{\ge}$

$ 1+\frac Rr+2\left(1+\frac rR\right)-3+5+\frac {2r}{R}=$ $ 5+\frac Rr+\frac {4r}{R}\ \Longrightarrow\ \sum\sqrt {\frac {b+c-a}{a}}\ \ge\ \sqrt {5+\frac Rr+\frac {4r}{R}}$ . Prove easily that the right inequality is truly. Indeed,

$ \sqrt {5+\frac Rr+\frac {4r}{R}}\ge\sqrt {9+2\cdot\left(1-\frac {2r}{R}\right)^2}\Longleftrightarrow$ $ 5+\frac Rr+\frac {4r}{R}\ge 9+2\cdot\left(1-\frac {2r}{R}\right)^2\Longleftrightarrow$ $ \frac {(R-2r)^2}{Rr}\ge 2\cdot\left(1-\frac {2r}{R}\right)^2\Longleftrightarrow (R-2r)^3\ge 0$ .
This post has been edited 7 times. Last edited by Virgil Nicula, Aug 14, 2017, 2:57 PM

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