247. An remarkable inequality with median.

by Virgil Nicula, Mar 6, 2011, 6:46 PM

$\boxed{\blacktriangleleft\ *\ \blacktriangleright}$ Research theme. Compare the ratios $\frac{m_{a}}{h_{a}}$ and $\frac{R}{2r}$ . If $\triangle ABC$ is $A$-right-angled or acute with $(a-b)(a-c)\ge 0$ , then prove easily that $\boxed{\ \frac {m_a}{h_a}\le\frac {R}{2r}\ }$ .

Indeed, $\frac{m_{a}}{h_{a}}\le \frac{R}{2r}\iff$ $am_a\le Rs$ . Since $m_a\le R(1+\cos A)=$ $2R\cos^2\frac A2=$ $\frac {2Rs(s-a)}{bc}$ obtain that $am_a\le \frac {2Ras(s-a)}{bc}$ . It is sufficiently

to prove $\frac {2Ras(s-a)}{bc}\le Rs$ , i.e.$2a(s-a)\le bc\iff$ $a(b+c-a)\le bc\iff$ $a^2-a(b+c)+bc\ge 0\iff$ $(a-b)(a-c)\ge 0$ , what is truly.


Remark. From the well-known identity $\boxed{\ \left(b^{2}-c^{2}\right)^{2}+16S^{2}=4a^{2}m^{2}_{a}\ }$ obtain that $\left(\frac{b^{2}-c^{2}}{4S}\right)^{2}+1=\left(\frac{2am_{a}}{4S}\right)^{2}$, i.e. $\left(\frac{b^{2}-c^{2}}{4S}\right)^{2}+1=\left(\frac{m_{a}}{h_{a}}\right)^{2}$. If

$m(\widehat{AMD})=\alpha$, then $\left\{\begin{array}{c}h_{a}=m_{a}\sin\alpha\\\\ \cot\alpha=\frac{|b^{2}-c^{2}|}{4S}\end{array}\right\|$ . Construct the parallelogram $ABA_{1}C\ .$ For a point $M$ in the triangle $ABC$ apply the Ptolemeu's inequality in the

quadrilateral $MBA_{1}C$ and we'll get $\left\{\begin{array}{c}c\cdot MB+b\cdot MC\ge BC\cdot MA_{1}\\\\ MA_{1}\ge 2m_{a}-MA\end{array}\right\|$ $\Longrightarrow$ $c\cdot MB+b\cdot MC\ge a(2m_{a}-MA)$ $\Longrightarrow$

$a\cdot MA+b\cdot MC+c\cdot MB\ge 2am_{a}$ . For $M:=O \implies$ $(a+b+c)R\ge 2am_{a}\Longleftrightarrow am_{a}\le pR\Longleftrightarrow 2S\frac{m_{a}}{h_{a}}\le 2S\frac{R}{2r}\Longleftrightarrow\boxed{\frac{m_{a}}{h_{a}}\le\frac{R}{2r}}$ .

Similarly $a\cdot MC+b\cdot MB+c\cdot MA\ge 2bm_{b};\ , a\cdot MB+b\cdot MA+c\cdot MC\ge 2cm_{c}$ $\implies$ $\boxed{MA+MB+MC\ge\frac{2\left(am_{a}+bm_{b}+cm_{c}\right)}{a+b+c}}$ .

For $M: =G\Longrightarrow $ $\boxed{(a+b+c)(m_{a}+m_{b}+m_{c})\ge 3(am_{a}+bm_{b}+cm_{c})}$ . For $M: =H\Longrightarrow \boxed{R+r\ge\frac{am_{a}+bm_{b}+cm_{c}}{a+b+c}}$ . Now you can

obtain interesting inequalities with these two inequalities or you can construct another remarkable quadrilateral $ABA_{2}C$ in which you"ll apply the Ptolemy's inequality.

I'll consider now the reflection $A'$ of $A$ w.r.t. the line $BC$ and I"ll apply the Ptolemeu's inequality to the convex quadrilateral $MBA'C$ :

$\left\{\begin{array}{c}b\cdot MB+c\cdot MC\ge a\cdot MA'\\\\ MA'\ge 2h_{a}-MA\end{array}\right\|$ $\Longrightarrow$ $\boxed{\ a\cdot MA+b\cdot MB+c\cdot MC\ge 4S\ }$, with equality iff $M: =H$, where $H$- the orthocenter of the triangle $ABC$.

Remark. For $M: =O\Longrightarrow$ $R\ge 2r$ and for $M: =I$ $\Longrightarrow$ $\boxed{\frac{p}{R}\le \sum \cos\frac{A}{2}}\le \frac{3\sqrt 3}{2}$.

Example. Apply the my general inequality to the Gergonne's triangle $A'B'C'$, where the incircle touches the sides of the triangle $ABC$ inscribed in the circle $w=C(O,R)$,

i.e $A'\in BC$ a.s.o. Thus, $S'=[A'B'C']=\frac{pr^{2}}{2R}$, $B'C'=2(p-a)\sin\frac{A}{2}$ a.s.o. and the my general inequality becomes $\sum (p-a)\sin\frac{A}{2}\cdot MA'\ge \frac{pr^{2}}{R}$ for

any inside point $M$. In particularly, for $M: =I$ obtain $\boxed{\ \sum (p-a)\sin\frac{A}{2}\ge \frac{pr}{R}\ }$ which is equivalently with $\sum\frac{p-a}{IA_{1}}\ge \frac{p}{R}$, where $A_{1}\in w\cap (AI$ a.s.o.

Remark of Mateescu Constantin. In the condition $(a-b)(a-c)\ge 0$ , the following stronger inequality holds : $\boxed{\ \frac {m_a}{h_a}\ \le\ \frac {\sqrt{a^2b^2+b^2c^2+c^2a^2}}{4S}\ }\ \le\ \frac R{2r}$ .

The boxed inequality holds if and only if $(a-b)(a-c)\ge 0$ . The second one is true in all triangles.
This post has been edited 49 times. Last edited by Virgil Nicula, Nov 22, 2015, 12:20 PM

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  • 391345 views moment

    by ryanbear, May 9, 2023, 6:10 AM

  • We need virgil nicula to return to aops, this blog is top 10 all time.

    by OlympusHero, Sep 14, 2022, 4:44 AM

  • :omighty: blog

    by tigerzhang, Aug 1, 2021, 12:02 AM

  • Amazing blog.

    by OlympusHero, May 13, 2021, 10:23 PM

  • the visits tho

    by GoogleNebula, Apr 14, 2021, 5:25 AM

  • Bro this blog is ripped

    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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