1989 OIM Problems/Problem 3

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

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1989 OIM Problems/Problem 3

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