Shortlist 2006/A5

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.

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