11. Inegalitatea I. Maftei & S. Radulescu (1).

by Virgil Nicula, Apr 20, 2010, 1:19 AM

Lema. In an acute-angled triangle $ ABC$ there is the inequality $ \boxed{\ \sin\frac A2\ + \ \sin\frac B2\ + \ \sin\frac C2\ \ge\ \frac 54\ + \ \frac r{2R}\ }$ .

Proof (Mateescu Constantin). Applying Popoviciu's inequality to the concave function $ \cos\ : \ \left(0\ ,\ \frac {\pi}{2}\right) \rightarrow (0,1)$ we get $ \cos A + \cos B + \cos C + 3\cos\frac {A + B + C}{3}\ \le$

$2\cos\frac {A + B}{2} +$ $ 2\cos\frac {B + C}{2} + 2\cos\frac {C + A}{2}$ $ \Longleftrightarrow\ 1 + \frac rR + \frac 32\ \le\ 2\sum\ \sin\frac A2\ \Longleftrightarrow\ \boxed{\ \sin\frac A2\ + \ \sin\frac B2\ + \ \sin\frac C2\ \ge\ \frac 54\ + \ \frac r{2R}\ }$ .

PP1. Prove that in the acute triangle $ ABC$ there is the following inequality $ \sum \sqrt {\frac {b + c - a}{a}}\ge 3$ .

Proof. Squaring both sides of the initial inequality we get $ \left(\sum\ \sqrt {\frac {b + c - a}a}\right)^2\ \ge\ 9\ \Longleftrightarrow$ $ \sum\ \frac {b + c - a}a + 2\sum\sqrt {\frac {(c + a - b)(a + b - c)}{bc}}\ \ge\ 9\ \Longleftrightarrow$

$ \sum\ \left(\frac ab + \frac ba\right) - 3 + 2\sum\sqrt {\frac {4(p - b)(p - c)}{bc}}\ \ge\ 9$ $ \Longleftrightarrow\ \sum\ \frac {a^2 + b^2}{ab} + 4\sum\ \sin\frac A2\ \ge\ 12$ . Taking into account the upper inequality it suffices to prove that

$ \sum\ \frac {a^2 + b^2}{ab} + 4\left(\frac 54 + \frac r{2R}\right)\ \ge\ 12\ \Longleftrightarrow$ $ \sum\ \frac {c^2 + 2ab\cos C}{ab} + \frac {2r}{R}\ \ge\ 7\ \Longleftrightarrow\ \sum\ \frac {c^2}{ab} + 2\sum\ \cos C + \frac {2r}R\ \ge\ 7$ $ \Longleftrightarrow\ \frac {a^3 + b^3 + c^3}{abc} + 2 + \frac {2r}{R} + \frac {2r}{R}\ \ge\ 7$ $ \Longleftrightarrow$

$\frac {2s(s^2 - 6Rr - 3r^2)}{4Rrs} + \frac {4r}{R}\ \ge\ 5\ \Longleftrightarrow\ \frac {s^2 - 6Rr - 3r^2}{2Rr} + \frac {4r}{R}\ \ge\ 5$ $ \Longleftrightarrow\ \frac {s^2 - 6Rr - 3r^2 + 8r^2}{2Rr}\ \ge\ 5\ \Longleftrightarrow\ s^2 + 5r^2\ \ge\ 16Rr$ , which is just Gerretsen's inequality.
This post has been edited 7 times. Last edited by Virgil Nicula, Nov 23, 2015, 7:17 AM

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