396. Own inegalities stronger than the Panaitopol's.

by Virgil Nicula, Jul 25, 2014, 2:02 AM

V.N.1 In $\triangle ABC$ denote $ 2s = a + b + c$ , the length $ r$ of inradius , the length $ R$ of circumradius and the length $ h_a$ of $ A$-altitude. Prove that there is the inequality

$ \boxed {R - 2r\ \ge\ \sqrt {s(s - a)} - h_a + \frac {r}{4a(s - a)}\cdot (b + c - 2a)^2}$ .Consequence. $ \boxed {R - 2r\ \ge\ \sqrt {s(s - a)} - h_a}$ (L.Panaitopol's inequality).

Proof. I will use the following trigonometrical identities : $\boxed{\ \begin{array}{cccc} 1\blacktriangleright & R-2r=R\left(1-4\sin\frac A2\cos\frac{B-C}2+4\sin^2\frac A2\right) \\ \\ 2\blacktriangleright & \sqrt{s(s-a)}=2R\cos\frac A2\cdot\sqrt{\cos^2\frac {B-C}2-\sin^2\frac A2} \\ \\ 3\blacktriangleright & h_a=2R\left(\cos^2\frac{B-C}2-\sin^2\frac A2\right) \\ \\ 4\blacktriangleright & \frac{r}{4a(s-a)}\cdot(b+c-2a)^2=R\left(\cos\frac{B-C}2-2\sin\frac A2 \right)^2\ \end{array} }$ . Therefore, the proposed inequality is equivalent to :

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

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

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 .


V.N.2 In an acute $\triangle ABC$ denote $ 2s = a + b + c$ , the length $ r$ of inradius , the length $ R$ of circumradius, the length $ h_a$ of $ A$-altitude and the length $ m_a$ of $ A$-median.

Prove that there the inequality $ \boxed {2(R - 2r)\ \ge\ m_a - h_a + \frac {r}{2a(s - a)}\cdot (b + c - 2a)^2}$ . Consequence. $ \boxed {2(R - 2r)\ \ge\ m_a - h_a}$ (L.Panaitopol's inequality).

Proof. I"ll show that, in $\triangle ABC$ satisfying $\angle A\le90^{\circ}$ the following inequality holds : $m_a-h_a\le 2R\sin^2\frac {B-C}{2}\ (\ast)$ . Indeed, since $\angle A\ \le\ 90^{\circ}\ \implies\ m_a\ \le\ R(1+\cos A)$ .

Thus, $m_a-h_a\ \le\ R(1+\cos A) - R\left(\cos A+\cos (B-C)\right)=2R\sin^2\frac {B-C}2$ . I used the relation (easy to prove) : $\boxed{\ h_a=R\left(\cos A+\cos (B-C)\right)\ }$ .

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 9 times. Last edited by Virgil Nicula, Jul 23, 2015, 2:56 AM

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    by samrocksnature, Apr 14, 2021, 5:16 AM

  • Holy- Darn this is good. shame it's inactive now

    by the_mathmagician, Jan 17, 2021, 7:43 PM

  • godly blog. opopop

    by OlympusHero, Dec 30, 2020, 6:08 PM

  • long blog

    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

    369302 views!

    -piphi

    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
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About Owner
  • Posts: 7054
  • Joined: Jun 22, 2005
Blog Stats
  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404395
  • Total comments: 37
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