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Difference between revisions of "1989 OIM Problems/Problem 3"
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== Solution ==
 
== Solution ==
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{{solution}}
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First, perform the Ravi Substitution; let <math>a=x+y,b=y+z,c=z+x</math> for positive <math>x,y,z</math>. Then the inequality becomes:
 +
<cmath>\iff\frac{x-z}{x+2y+z}+\frac{y-x}{x+y+2z}+\frac{z-y}{2x+y+z}<\frac{1}{16}</cmath>
 +
<cmath>\iff\frac{2x+2y}{x+2y+z}-1+\frac{2y+2z}{x+y+2z}-1+\frac{2z+2x}{2x+y+z}-1<\frac{1}{16}</cmath>
 +
<cmath>\iff\frac{x+y}{x+2y+z}+\frac{y+z}{x+y+2z}+\frac{z+x}{2x+y+z}<\frac{49}{32}</cmath>
 +
Next, since the inequality is homogenized, assume without loss of generality that <math>x+y+z=1</math>; then,
 +
<cmath>\iff\frac{1-z}{y+1}+\frac{1-x}{z+1}+\frac{1-y}{x+1}<\frac{49}{32}</cmath>
 +
Instead of proving the above, we prove a stronger inequality:
 +
<cmath>\frac{1-z}{y+1}+\frac{1-x}{z+1}+\frac{1-y}{x+1}\le\frac{3}{2}</cmath>
 +
Multiplying out both sides:
 +
<cmath>\iff(1-z)(z+1)(x+1)+(1-x)(x+1)(y+1)+(1-y)(y+1)(z+1)\le\frac{3}{2}(x+1)(y+1)(z+1)</cmath>
 +
<cmath>\iff(1-z^2)(x+1)+(1-x^2)(y+1)+(1-y^2)(z+1)\le\frac{3}{2}(xyz+xy+yz+zx+(x+y+z)+1)</cmath>
 +
<cmath>\iff3+(x+y+z)-(x^2+y^2+z^2)-(x^2y+y^2z+z^2x)\le\frac{3}{2}(xyz+xy+yz+zx+2)</cmath>
 +
<cmath>\iff8-2(x^2+y^2+z^2)-2(x^2y+y^2z+z^2x)\le3xyz+3(xy+yz+zx)+6</cmath>
 +
<cmath>\iff3xyz+3(xy+yz+zx)+2(x^2+y^2+z^2)+2(x^2y+y^2z+z^2x)\ge 2</cmath>
 +
<cmath>\iff3xyz+2(x+y+z)^2-(xy+yz+zx)+2(x^2y+y^2z+z^2x)\ge 2</cmath>
 +
<cmath>\iff3xyz-(xy+yz+zx)+2(x^2y+y^2z+z^2x)\ge 0</cmath>
 +
Next, assume without loss of generality that <math>x\ge y\ge z>0</math>. Since <math>x+y+z=1</math>, we must have <math>\frac{1}{3}\le x<1</math>. The upper bound is obvious (recall that <math>y</math> and <math>z</math> are positive), so we show the lower bound. Assume for the sake of contradiction that <math>x<\frac{1}{3}</math>; then,
 +
<cmath>x+y+z\le x+x+x=3x<3\cdot\frac{1}{3}=1</cmath>
 +
implying <math>x+y+z<1</math>, a contradiction, so the bounds hold. Then,
 +
<cmath>\iff3xyz-(x(y+z)+yz)+2(x^2y+y^2z+z^2x)\ge 0</cmath>
 +
<cmath>\iff yz(3x-1)-x(1-x)+2(x^2y+y^2z+z^2x)\ge 0</cmath>
 +
<cmath>\iff yz(3x-1)+2(x^2y+y^2z+z^2x)\ge x(1-x)</cmath>
 +
Due to the bounds, the left-hand side is nonnegative, but the right-hand side is nonpositive, so we are done.
 +
 
 +
~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
  
 
== See also ==
 
== See also ==
 
https://www.oma.org.ar/enunciados/ibe4.htm
 
https://www.oma.org.ar/enunciados/ibe4.htm

Latest revision as of 20:01, 19 April 2025

Problem

Let $a$, $b$, and $c$ be the longitudes of the sides of a triangle. Prove: \[\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}<\frac{1}{16}\]

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

First, perform the Ravi Substitution; let $a=x+y,b=y+z,c=z+x$ for positive $x,y,z$. Then the inequality becomes: \[\iff\frac{x-z}{x+2y+z}+\frac{y-x}{x+y+2z}+\frac{z-y}{2x+y+z}<\frac{1}{16}\] \[\iff\frac{2x+2y}{x+2y+z}-1+\frac{2y+2z}{x+y+2z}-1+\frac{2z+2x}{2x+y+z}-1<\frac{1}{16}\] \[\iff\frac{x+y}{x+2y+z}+\frac{y+z}{x+y+2z}+\frac{z+x}{2x+y+z}<\frac{49}{32}\] Next, since the inequality is homogenized, assume without loss of generality that $x+y+z=1$; then, \[\iff\frac{1-z}{y+1}+\frac{1-x}{z+1}+\frac{1-y}{x+1}<\frac{49}{32}\] Instead of proving the above, we prove a stronger inequality: \[\frac{1-z}{y+1}+\frac{1-x}{z+1}+\frac{1-y}{x+1}\le\frac{3}{2}\] Multiplying out both sides: \[\iff(1-z)(z+1)(x+1)+(1-x)(x+1)(y+1)+(1-y)(y+1)(z+1)\le\frac{3}{2}(x+1)(y+1)(z+1)\] \[\iff(1-z^2)(x+1)+(1-x^2)(y+1)+(1-y^2)(z+1)\le\frac{3}{2}(xyz+xy+yz+zx+(x+y+z)+1)\] \[\iff3+(x+y+z)-(x^2+y^2+z^2)-(x^2y+y^2z+z^2x)\le\frac{3}{2}(xyz+xy+yz+zx+2)\] \[\iff8-2(x^2+y^2+z^2)-2(x^2y+y^2z+z^2x)\le3xyz+3(xy+yz+zx)+6\] \[\iff3xyz+3(xy+yz+zx)+2(x^2+y^2+z^2)+2(x^2y+y^2z+z^2x)\ge 2\] \[\iff3xyz+2(x+y+z)^2-(xy+yz+zx)+2(x^2y+y^2z+z^2x)\ge 2\] \[\iff3xyz-(xy+yz+zx)+2(x^2y+y^2z+z^2x)\ge 0\] Next, assume without loss of generality that $x\ge y\ge z>0$. Since $x+y+z=1$, we must have $\frac{1}{3}\le x<1$. The upper bound is obvious (recall that $y$ and $z$ are positive), so we show the lower bound. Assume for the sake of contradiction that $x<\frac{1}{3}$; then, \[x+y+z\le x+x+x=3x<3\cdot\frac{1}{3}=1\] implying $x+y+z<1$, a contradiction, so the bounds hold. Then, \[\iff3xyz-(x(y+z)+yz)+2(x^2y+y^2z+z^2x)\ge 0\] \[\iff yz(3x-1)-x(1-x)+2(x^2y+y^2z+z^2x)\ge 0\] \[\iff yz(3x-1)+2(x^2y+y^2z+z^2x)\ge x(1-x)\] Due to the bounds, the left-hand side is nonnegative, but the right-hand side is nonpositive, so we are done.

~ eevee9406

See also

https://www.oma.org.ar/enunciados/ibe4.htm

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