Shortlist 2006/A5

By giuricrocodilePosted on December 22, 2020January 7, 2021Tagged Algebra, Inequality, Substitution

Let $a,b,c$ be the sides of a triangle. Prove that

$\sum_{cyc}\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}} \leq 3$

Always Try A Substitution

We notice that the denominators are all positive, in fact $\sqrt{a}+\sqrt{b}>\sqrt{a+b}>\sqrt{c}$ by the triangular inequality. So let $x=\sqrt{b}+\sqrt{c}-\sqrt{a}, y=\sqrt{a}+\sqrt{b}-\sqrt{c}, z= \sqrt{c}+\sqrt{a}-\sqrt{b}$ . Then $b+c-a= (\frac{z+x}{2})^2+(\frac{x+y}{2})^2-(\frac{y+z}{2})^2=x^2-\frac{1}{2}(x-y)(x-z)$ . So we get

$\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}} = \sqrt{1-\frac{(x-y)(x-z)}{2x^2}} \leq 1-\frac{(x-y)(x-z)}{4x^2}$

applying $\sqrt{1+2u} \leq 1+u$ in the last step. Similarly, we obtain

$\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}} \leq 1-\frac{(z-x)(z-y)}{4z^2}$

$\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}} \leq 1-\frac{(y-x)(y-z)}{4y^2}$

Thus, now it is sufficient to prove that

$\frac{(x-y)(x-z)}{x^2}+\frac{(y-x)(y-z)}{y^2}+\frac{(z-x)(z-y)}{z^2} \geq 0$

So let’s assume WLOG $x \leq y \leq z$ , then we have

$\frac{(y-x)(z-x)}{x^2} \geq \frac{(y-x)(z-y)}{y^2}=-\frac{(y-x)(y-z)}{y^2}$

$\frac{(z-x)(z-y)}{z^2} \geq 0$

Hence, we are done.

Note

$\frac{(x-y)(x-z)}{x^2}+\frac{(y-x)(y-z)}{y^2}+\frac{(z-x)(z-y)}{z^2} \geq 0$

is a special case of the well-known inequality

$x^t(x-y)(x-z)+y^t(y-x)(y-z)+z^t(z-x)(z-y)\geq 0$

whose case with $t=0$ is known as Schur’s Inequality.