15. Inequalities stronger than Panaitopol's.

by Virgil Nicula, Apr 20, 2010, 1:45 AM

PP (two inequalities). Let $\triangle ABC$ with the semiperimeter $s$ and the lengths $\left(r,R,h_a,m_a\right)$ of inradius, circumradius, $A$-altitude

and $A$-median. Prove that there is the inequalities $:\ \left\{\begin{array}{cccc}
R - 2r & \ge & \sqrt {s(s - a)} - h_a + \frac {r}{4a(s - a)}\cdot (b + c - 2a)^2 & (1)\\\\
2(R - 2r) & \ge & m_a - h_a + \frac {r}{2a(s - a)}\cdot (b + c - 2a)^2 & (2)\end{array}\right\|$

Consequencies.
$ \boxed {R - 2r\ \ge\ \sqrt {s(s - a)} - h_a}$ (L.Panaitopol's inequality) and $\boxed{2(R - 2r)\ \ge\ m_a - h_a}$ (L.Panaitopol's inequality).

Proof 1.1(own - with out trigonometry).

Proof 1.2 (own - without trigonometry).

Proof 2.1 (Mateescu Constantin). I"ll use $:\ \boxed{\begin{array}{cc}
(1) & R-2r=R\left(1-4\sin\frac A2\cos\frac{B-C}2+4\sin^2\frac A2\right) \\ 
 \\ 
(2) & \sqrt{s(s-a)}=2R\cos\frac A2\cdot\sqrt{\cos^2\frac {B-C}2-\sin^2\frac A2}\end{array}}$ and $\boxed{\begin{array}{cc}
(3) & h_a=2R\left(\cos^2\frac{B-C}2-\sin^2\frac A2\right) \\ 
 \\ 
(4) & \frac{r}{4a(s-a)}\cdot(b+c-2a)^2=R\left(\cos\frac{B-C}2-2\sin\frac A2 \right)^2\ \end{array}}\ .$

Inequality becomes $:\ 1-4\sin\frac A2\cos\frac {B-C}2+4\sin^2\frac A2\ge$ $2\cos\frac A2\sqrt{\cos^2\frac {B-C}2-\sin^2\frac A2}-$ $2\left(\cos^2\frac {B-C}2-\sin^2\frac A2\right)+$ $\left(\cos\frac {B-C}2-2\sin\frac A2\right)^2\iff$

$1+\cos^2\frac {B-C}2-2\sin^2\frac A2\ge$ $2\cos\frac A2\sqrt{\cos^2\frac {B-C}2-\sin^2\frac A2}\iff$ $\cos^2\frac A2+\left(\cos^2\frac {B-C}2-\sin^2\frac A2\right)\ge$ $2\cos\frac A2\sqrt{\cos^2\frac {B-C}2-\sin^2\frac A2}\ .$ Now, the last inequality

follows directly from the well-known AM-GM inequality . The expression under the square root is positive according to identity 3 .

Proof 2.2 Mateescu Constantin). Firstly we will show that, in a triangle $ABC$ satisfying $\angle A\ \le\ 90^{\circ}$ the following inequality holds : $\boxed{\ m_a-h_a\ \le\ 2R\sin^2\frac {B-C}{2}\ }\ \ (\ast)$ . Proof

Now I'll use the identities $:\ \boxed{\ \begin{array}{cccc}
1\blacktriangleright & R-2r=R\left(1-4\sin\frac{A}{2}\cos\frac{B-C}{2}+4\sin^{2}\frac{A}{2}\right)\\ \\ 
2\blacktriangleright &\frac{r}{2a(s-a)}\cdot(b+c-2a)^{2}=2R\left(\cos\frac{B-C}{2}-2\sin\frac{A}{2}\right)^{2}\ \end{array}} $ and the upper inequality$.$ Thus, $m_a-h_a+\frac{r}{2a(s-a)}\cdot (b+c-2a)^2\ \stackrel{\ast\wedge 2}{\le}$

$2R\sin^2\frac {B-C}2$ $+2R\left(\cos\frac {B-C}2-2\sin\frac A2\right)^2$ $\iff$ $m_a-h_a+\frac{r}{2a(s-a)}\cdot (b+c-2a)^2\ \le\ 2R$ $\left(1-4\cos\frac{B-C}2\sin\frac A2+4\sin^2\frac A2\right)\stackrel{(1)}{=}2(R-2r)$ .
This post has been edited 33 times. Last edited by Virgil Nicula, Nov 4, 2016, 2:41 PM

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