Geometric inequality!

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Geometric inequality!

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Source: India TST 2001 Day 1 Problem 3

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#1 h Jan 31, 2015, 7:51 AM • 2 Y

Y by Adventure10, Mango247

In a triangle $ABC$ with incircle $\omega$ and incenter $I$ , the segments $AI$ , $BI$ , $CI$ cut $\omega$ at $D$ , $E$ , $F$ , respectively. Rays $AI$ , $BI$ , $CI$ meet the sides $BC$ , $CA$ , $AB$ at $L$ , $M$ , $N$ respectively. Prove that:
$AL+BM+CN \leq 3(AD+BE+CF)$
When does equality occur?

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#2 h Mar 23, 2018, 1:46 PM • 2 Y

Y by Adventure10, Mango247

can you prove it

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#3 h Jan 29, 2019, 2:13 AM • 2 Y

Y by Adventure10, Mango247

consider $f(x)=csc(x),f''(x)>0$

This post has been edited 1 time. Last edited by Untro368, May 7, 2020, 12:14 AM
Reason: typo

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Mateescu Constantin

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Mateescu Constantin

#4 h Jan 29, 2019, 1:17 PM • 3 Y

Y by TFIRSTMGMEDALIST, Adventure10, Mango247

This is a super weak inequality... In terms of classical triangle elements, the conclusion is equivalent to:

$l_a + l_b + l_c \le 3(IA + IB + IC) - 9r$
$\iff 9r + 2(l_a + l_b + l_c) \le (l_a + l_b + l_c) \le 3 (IA + IB + IC)$
which follows from:

$\boxed{9r \stackrel{\text{(Jensen)}}{\le} 3r + (IA + IB + IC) \le l_a + l_b + l_c}$
and

$\boxed{ 2(l_a + l_b + l_c) \le 3(PA + PB + PC) }\ ,$
where $P$ is an arbitrary point in the plane of $\triangle ABC$ (see here).

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MathPassionForever

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MathPassionForever

#5 h Mar 12, 2021, 3:35 AM • 2 Y

Y by Pluto1708, Mango247

We assume standard notation, except $r$ is not the inradius. We consider the sharper inequality

$3(AI + BI + CI) \ge 2(AD + BE + CF)$
where $D,E,F$ play the role of $L,M,N$ in the original problem.

We note that

$\dfrac{AI}{AD} = \dfrac{b+c}{2s}$
Defining $p,q,r,x,y,z$ such that $AD = 2sp, BE = 2sq, CF = 2sr, a=y+z, b=z+x, c=x+y$ , we have

$AI = (x+s)p, BI = (y+s)q, CI = (z+s)r$
The original inequality becomes

$3\sum (2x+y+z)p \ge 2 \sum 2(x+y+z)p \iff \sum (2x - y - z)p \ge 0 \iff \sum (p-q)(x-y) \ge 0$
Each term in the rightmost side is nonnegative since it's well-known that if $b > c \iff z > y$ , then $q > r$ .

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#6 h Mar 14, 2021, 2:14 PM

Y by

ComplexPhi wrote:

In a triangle $ABC$ with incircle $\omega$ and incenter $I$ , the segments $AI$ , $BI$ , $CI$ cut $\omega$ at $D$ , $E$ , $F$ , respectively. Rays $AI$ , $BI$ , $CI$ meet the sides $BC$ , $CA$ , $AB$ at $L$ , $M$ , $N$ respectively. Prove that:
$AL+BM+CN \leq 3(AD+BE+CF)$ When does equality occur?

Using Ravi's substitution the inequality is equivalent to $\sum \left(\dfrac{4x+y+z}{2x+y+z}\cdot \sqrt{\dfrac{(x+y)(x+z)}{yz}}\right)\geq 9$ WLOG let $x+y+z=1$ and define functions $a(x)=\dfrac{3x+1}{x+1}\; b(x)=\sqrt{\dfrac{1-x}{x}}\; f(x)=\tfrac{a(x)}{b(x)}$
Next we show $f(x)\geq \tfrac{9x}{2\sqrt{2}}$ .Indeed this is equivalent to $8x(3x+1)^2\geq 81x(1-x)(1+x)^2=(3x-1)^2(9x^2+15x+8)\geq 0$ Hence we have $\dfrac{a(x)}{b(x)}+\dfrac{a(y)}{b(y)}+\dfrac{a(z)}{b(z)}=f(x)+f(y)+f(z)\geq \tfrac{9}{2\sqrt{2}}\geq \dfrac{9}{b(x)b(y)b(z)}$ Hence $a(x)b(x)+a(y)b(y)+a(z)b(z)\geq 9$ but clearly this what we originally wanted. $\square$

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