IMO 2006 Shortlisted Problems

Note first that the denominators are all positive, e.g. $\sqrt{a}+\sqrt{b}>\sqrt{a+b}>\sqrt{c}$. Let $x=\sqrt{b}+\sqrt{c}-\sqrt{a}$, $y=\sqrt{c}+\sqrt{a}-\sqrt{b}$ and $z=\sqrt{a}+\sqrt{b}-\sqrt{c}$. Then $b+c-a={\left(\frac{z+x}{2}\right)}^2+{\left(\frac{x+y}{2}\right)}^2-{\left(\frac{y+z}{2}\right)}^2=\frac{x^2+xy+xz-yz}{2}=x^2-\frac{1}{2}(x-y)(x-z)$ and $$\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}=\sqrt{1-\frac{(x-y)(x-z)}{2x^2}}\le 1-\frac{(x-y)(x-z)}{4x^2},$$ applying $\sqrt{1+2u}\le 1+u$ in the last step. Similarly we obtain $$\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}\le 1-\frac{(z-x)(z-y)}{4z^2}\text{ and }\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\le 1-\frac{(y-z)(y-x)}{4y^2}.$$ Substituting these quantities into the statement, it is sufficient to prove that $$\begin{array}{c}\frac{(x-y)(x-z)}{x^2}+\frac{(y-z)(y-x)}{y^2}+\frac{(z-x)(z-y)}{z^2}\ge 0.\end{array}\text{\hspace{1em}(1)}$$ By symmetry we can assume $x\le y\le z$. Then $$\begin{array}{c}\frac{(x-y)(x-z)}{x^2}=\frac{(y-x)(z-x)}{x^2}\ge \frac{(y-x)(z-y)}{y^2}=-\frac{(y-z)(y-x)}{y^2}\\ \frac{(z-x)(z-y)}{z^2}\ge 0\end{array}$$ and (1) follows.

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Alternative solution.

Due to the symmetry of variables, it can be assumed that $a\ge b\ge c$. We claim that $$\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\le 1\text{ and }\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}\le 2.$$ The first inequality follows from $$\sqrt{a+b-c}-\sqrt{a}=\frac{(a+b-c)-a}{\sqrt{a+b-c}+\sqrt{a}}\le \frac{b-c}{\sqrt{b}+\sqrt{c}}=\sqrt{b}-\sqrt{c}.$$ For proving the second inequality, let $p=\sqrt{a}+\sqrt{b}$ and $q=\sqrt{a}-\sqrt{b}$. Then $a-b=pq$ and the inequality becomes $$\frac{\sqrt{c-pq}}{\sqrt{c}-q}+\frac{\sqrt{c+pq}}{\sqrt{c}+q}\le 2.$$ From $a\ge b\ge c$ we have $p\ge 2\sqrt{c}$. Applying the Cauchy-Schwarz inequality, $$\begin{array}{c}{\left(\frac{\sqrt{c-pq}}{\sqrt{c}-q}+\frac{\sqrt{c+pq}}{\sqrt{c}+q}\right)}^2\le \left(\frac{c-pq}{\sqrt{c}-q}+\frac{c+pq}{\sqrt{c}+q}\right)\left(\frac{1}{\sqrt{c}-q}+\frac{1}{\sqrt{c}+q}\right)\\ =\frac{2\left(c\sqrt{c}-p{q}^2\right)}{c-q^2}\cdot \frac{2\sqrt{c}}{c-q^2}=4\cdot \frac{c^2-\sqrt{c}p{q}^2}{{\left(c-q^2\right)}^2}\le 4\cdot \frac{c^2-2c{q}^2}{{\left(c-q^2\right)}^2}\le 4\end{array}$$