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wer

#1 h Jul 5, 2014, 4:46 AM • 3 Y

Y by Adventure10, Mango247, qwedsazxc

If a, b, c are the sides of a triangle, show that: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{r}{R}\le2$

This post has been edited 1 time. Last edited by wer, Feb 23, 2016, 2:19 PM
Reason: ................

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arqady

#2 h Jul 5, 2014, 5:45 AM • 2 Y

Y by Adventure10, Mango247

wer wrote:

If a, b, c are the sides of a triangle, show that: $a/(b+c)+b/(c+a)+c/(a+b)+r/R\leq2$ .

After Ravi's substitution and full expanding we need to prove that
$\sum_{sym}(3x^4y^2-x^4yz+3x^3y^3-2x^3y^2z-3x^2y^2z^2)\geq0$ , which is obviously true.

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42171 posts

sqing

#3 h Jul 5, 2014, 7:40 AM • 2 Y

Y by Adventure10, Mango247

If $a, b, c$ are the sides of a triangle, show that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{r}{R}\leq2.$
Outstanding inequality.

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39 posts

#4 h Dec 28, 2014, 5:19 PM • 1 Y

Y by Adventure10

$nice\: \: \: \: inequality$

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1269 posts

wer

#5 h Feb 23, 2016, 2:49 PM • 3 Y

Y by Filipjack, Adventure10, Mango247

my solution(from 2014) After Ravi's subtitution, we need $\frac{x+y}{2z+x+y}+\frac{y+z}{2x+y+z}+\frac{y+z}{2x+y+z}+\frac{4xyz}{(x+y)(y+z)(z+x)}\le2$ .
Further $\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+z)(y+x)}\ge\frac{x+y}{x+y+2z}$ and analogus,
so $2(\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+z)(y+x)}+\frac{z^2}{(z+x)(z+y)})\ge \frac{x+y}{x+y+2z}+\frac{y+z}{y+z+2x}+\frac{z+x}{z+x+2y}$ ,result
$2(\frac{x^2y+x^2z+y^2z+y^2x+z^2x+z^2y}{(x+y)(y+z)(z+x)})\ge\frac{x+y}{x+y+2z}+\frac{y+z}{y+z+2x}+\frac{z+x}{z+x+2y}$ , result
$2\frac{(x+y)(y+z)(z+x)-2xyz}{((x+y)(y+z)(z+x)}\ge\frac{x+y}{x+y+2z}+\frac{y+z}{y+z+2x}+\frac{z+x}{z+x+2y}$ , so
$\frac{x+y}{2z+x+y}+\frac{y+z}{2x+y+z}+\frac{y+z}{2x+y+z}+\frac{4xyz}{(x+y)(y+z)(z+x)}\le2$ .

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#6 h Feb 23, 2016, 4:04 PM • 2 Y

Y by Adventure10, Mango247

Dear wer
Very nice solution.

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42171 posts

sqing

#7 h Feb 23, 2016, 10:44 PM • 1 Y

Y by Adventure10

Thank wer
Very nice solution.
here

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sqing

#8 h Dec 18, 2016, 8:11 AM • 1 Y

Y by Adventure10

Mathematical Gazette nr. 11/2016, problem 27298
http://www.ssmrmh.ro/wp-content/uploads/2016/12/COMMENTED-PROBLEM-3.pdf

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sqing

#9 h Mar 19, 2018, 2:13 PM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

If $a, b, c$ are the sides of a triangle, show that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{r}{R}\leq2.$ Outstanding inequality.

If $a, b, c$ are the sides of a triangle, show that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq2-\frac{r}{R}\leq\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2}.$ #8

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xzlbq

#11 h Mar 20, 2018, 12:12 AM • 2 Y

Y by Adventure10, Mango247

$2\, \left( {\frac {{\it wb}\,{\it wc}}{{\it wa}}}+{\frac {{\it wc}\,{ \it wa}}{{\it wb}}}+{\frac {{\it wa}\,{\it wb}}{{\it wc}}} \right) \left( {\it ra}+{\it rb}+{\it rc} \right) ^{-1}-{\frac {r}{R}}\geq{ \frac {a}{b+c}}+{\frac {b}{c+a}}+{\frac {c}{a+b}}$

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15849 posts

xzlbq

#12 h Mar 20, 2018, 12:20 AM • 2 Y

Y by Adventure10, Mango247

$2-\frac{3}{2}\,{\frac {\sqrt {3}r}{s}}\geq {\frac {a}{b+c}}+{\frac {b}{c+a}}+{ \frac {c}{a+b}}$

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15849 posts

xzlbq

#13 h Mar 20, 2018, 12:28 AM • 2 Y

Y by Adventure10, Mango247

$2-\frac{3}{2}\,{\frac {\sqrt {3}r}{s}}-{\frac {a}{b+c}}-{\frac {b}{c+a}}-{ \frac {c}{a+b}}=\frac{1}{6}\,{\frac {rQ\sqrt {3}}{{s}^{2} \left( {s}^{2}+r \left( r+2\,R \right) \right) }}$

$Q=\left( R-2\,r \right) ^{2} \left( 192\,\sqrt {3}-324 \right) r+ \left( R-2\,r \right) \left( {s}^{2}-16\,Rr+5\,{r}^{2} \right) \left( -18+12\,\sqrt {3} \right) +162\, \left( R-2\,r \right) \left( 2\,R-4\,r-s+3\,\sqrt {3}r \right) r+9\, \left( {s}^{2}-16\,Rr+ 5\,{r}^{2} \right) \left( 2\,R-4\,r-s+3\,\sqrt {3}r \right) + \left( R-2\,r \right) \left( 350\,\sqrt {3}-576 \right) {r}^{2}+5\, \left( { s}^{2}-16\,Rr+5\,{r}^{2} \right) \sqrt {3}r+288\, \left( 2\,R-4\,r-s+3 \,\sqrt {3}r \right) {r}^{2}$

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Desimathematics

465 posts

Desimathematics

#14 h Mar 20, 2018, 3:42 AM • 2 Y

Y by Adventure10, Mango247

We know that (a/b+c)<=1/2 similarly
(b/a+c)<=1/2 and also (c/a+b)<=1/2
We also know that r/R<=1/2 adding all the these relations we get the given beautiful inequality

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42171 posts

sqing

#15 h Apr 11, 2025, 2:33 AM

Y by

wer wrote:

If a, b, c are the sides of a triangle, show that: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{r}{R}\le2$

Mathematics Magazine Volume 97 (2024):A geometric inequality October 2023
Prove that in any triangle with side lengths $a,b,c$ inradius $r$ , and circumradius $R$ , we have
$\frac{a}{b+c}+ \frac{b}{c+a}+ \frac{c}{a+b}+ \frac{r}{R} > \frac{5}{3}$ Problem 2077 in the October 2019 issue of Mathematics Magazine

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#16 h Apr 11, 2025, 2:49 AM

Y by

Desimathematics wrote:

We know that (a/b+c)<=1/2 similarly
(b/a+c)<=1/2 and also (c/a+b)<=1/2
We also know that r/R<=1/2 adding all the these relations we get the given beautiful inequality

this is really, really nice. i thought the only thing about this related to geo was ravi substitution. Euler inequality was pretty unexpected

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481 posts

#17 h Apr 11, 2025, 4:20 AM

Y by

But unfortunately, this is wrong. For example, a $13-14-15$ triangle with $a=15, b=14, c=13$ will yield $\frac{a}{b+c} = \frac{15}{27} > \frac{1}{2}.$

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#18 h Apr 11, 2025, 12:25 PM

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oh... hold on a second. $\frac{r}{R} \leq \frac{1}{2}$ is true right? So the result follows from Nesbitt's inequality. I suppose the triangle sidelengths don't really matter here?

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arqady

#19 h Apr 11, 2025, 3:23 PM

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mpcnotnpc wrote:

oh... hold on a second. $\frac{r}{R} \leq \frac{1}{2}$ is true right? So the result follows from Nesbitt's inequality. I suppose the triangle sidelengths don't really matter here?

Are you sure?

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481 posts

#20 h Apr 12, 2025, 12:10 AM

Y by

mpcnotnpc wrote:

oh... hold on a second. $\frac{r}{R} \leq \frac{1}{2}$ is true right? So the result follows from Nesbitt's inequality. I suppose the triangle sidelengths don't really matter here?

$\sum_{cyc} \frac{a}{b+c} \geq \frac{3}{2}.$ $r/R \leq \frac{1}{2}.$ $\sum_{cyc} \frac{a}{b+c} +\frac{r}{R} \quad ??? \quad 2$

This post has been edited 1 time. Last edited by anduran, Apr 12, 2025, 12:10 AM

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#21 h Apr 12, 2025, 2:16 AM

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oh bruh :sob:

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