Prove that $\sum_{cyc}\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\le 3$

Question

Let $a,b$ and $c$ be the sides of a triangle. Prove that $$\sum_{cyc}\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\le 3$$

This is from the IMO shortlist the year 2006. I'm not asking for a solution, I just want to know how to finish my approach,

Using Cauchy-Shwartz, it is enough to prove $$\left(\sum_{cyc}\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\right)^2\le 3\sum_{cyc}\frac{a+b-c}{(\sqrt{a}+\sqrt{b}-\sqrt{c})^2}\le 9$$ or equivalently $$\sum_{cyc}\frac{a+b-c}{(\sqrt{a}+\sqrt{b}-\sqrt{c})^2}\le 3$$ Let $a=x^2,b=y^2$ and $c=z^2$ and assume wlog $\sqrt{a}+\sqrt{b}-\sqrt{c}=x+y-z=1$ We want to show $$\sum_{cyc}\frac{x^2+y^2-z^2}{1}=x^2+y^2+z^2\le 3$$

But $x+y=1+z$ meaning $x^2+y^2+2xy=z^2+2z+1 \implies x^2+y^2+z^2=2z^2+2z+1-2xy.$ So we want to show $$z^2+z\le 1+xy$$ But how can I prove this?

Answer 1

I have a different method, if you are interested.
Let $x=\sqrt a$, $y=\sqrt b$ and $z=\sqrt c$, which are sides of a triangle. $$\sum_{cyc}\frac{\sqrt{a+b-c}}{\sqrt a-\sqrt b-\sqrt c}=\sum_{cyc}\frac{\sqrt{x^2+y^2-z^2}}{x+y-z}$$ $$\cos Z=\frac{x^2+y^2-z^2}{2xy}\Rightarrow \sqrt{x^2+y^2-z^2}=\sqrt{2xy\cos Z}$$ $$\therefore\sum_{cyc}\frac{\sqrt{x^2+y^2-z^2}}{x+y-z}=\sum_{cyc}\frac{\sqrt{2xy\cos Z}}{x+y-z}$$ By the weighted AM-GM inequality: $$\sum_{cyc}\frac{\sqrt{2xy\cos Z}}{x+y-z}\le\sum_{cyc}\left(\frac{x+y}{2}+2\cos Z-x-y+z\right)=2\sum_{cyc}\cos X\le3$$

Answer 2

Let $a\geq b\geq c$.

Thus, $$3-\sum_{cyc}\frac{\sqrt{a+b-c}}{\sqrt a+\sqrt b-\sqrt c}=\sum_{cyc}\left(1-\frac{\sqrt{a+b-c}}{\sqrt a+\sqrt b-\sqrt c}\right)=$$ $$=\sum_{cyc}\frac{\sqrt{a}+\sqrt{b}-\sqrt{c}-\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}=$$ $$=2\sum_{cyc}\frac{\sqrt{ab}-\sqrt{c(a+b-c)}}{\left(\sqrt a+\sqrt b-\sqrt c\right)\left(\sqrt a+\sqrt b+\sqrt c+\sqrt{a+b-c}\right)}=$$ $$=2\sum_{cyc}\frac{(a-b)(a-c)}{(\sqrt{bc}+\sqrt{a(b+c-a)})(\sqrt b+\sqrt c-\sqrt a)\left(\sqrt a+\sqrt b+\sqrt c+\sqrt{b+c-a}\right)}\geq$$ $$\geq\frac{2(a-b)(a-c)}{(\sqrt{bc}+\sqrt{a(b+c-a)})(\sqrt b+\sqrt c-\sqrt a)\left(\sqrt a+\sqrt b+\sqrt c+\sqrt{b+c-a}\right)}+$$ $$+\frac{2(b-a)(b-c)}{(\sqrt{ac}+\sqrt{b(a+c-b)})(\sqrt a+\sqrt c-\sqrt b)\left(\sqrt a+\sqrt b+\sqrt c+\sqrt{a+c-b}\right)}=$$ $$=2(a-b)\left(\tfrac{a-c}{(\sqrt{bc}+\sqrt{a(b+c-a)})(\sqrt b+\sqrt c-\sqrt a)\left(\sqrt a+\sqrt b+\sqrt c+\sqrt{b+c-a}\right)}-\tfrac{b-c}{\left(\sqrt{ac}+\sqrt{b(a+c-b)}\right))(\sqrt a+\sqrt c-\sqrt b)\left(\sqrt a+\sqrt b+\sqrt c+\sqrt{a+c-b}\right)}\right)\geq0$$ because $$a-c\geq b-c,$$ $$\frac{1}{\sqrt{bc}+\sqrt{a(b+c-a)}}\geq\frac{1}{\sqrt{ac}+\sqrt{b(a+c-b)}},$$ $$\frac{1}{\sqrt b+\sqrt c-\sqrt a}\geq\frac{1}{\sqrt a+\sqrt c-\sqrt b}$$ and $$\frac{1}{\sqrt a+\sqrt b+\sqrt c+\sqrt{b+c-a}}\geq\frac{1}{\sqrt a+\sqrt b+\sqrt c+\sqrt{a+c-b}}.$$

Answer 3

I'll start with the step in which you found that we need to prove $$\sum_{cyc}\frac{a+b-c}{\left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)^2}\le 3.$$ Your replacement did solve the square roots, but did not make the most of the condition that $a$, $b$ and $c$ are the sides of a triangle. Notice that $a+b+2\sqrt{ab}>a+b>c$, so $\sqrt a+\sqrt b>\sqrt c$. Similarly, we can prove that $\sqrt a$, $\sqrt b$ and $\sqrt c$ are the sides of a triangle as well.

Therefore, let $a=(m+n)^2$, $b=(n+p)^2$ and $c=(p+m)^2$. Plug in and numden, we need to prove \begin{align*}&\sum\frac{2m^2+2mn+2mp-2np}{4m^2}=\frac{m^2+mn+mp-np}{2m^2}\le3\\={}&\frac{3m^2n^2p^2+\sum\left(m^3n^2p+m^3np^2\right)-\sum m^3n^3}{2m^2n^2p^2}\overset*\le3.\end{align*} Here, $*$ is simply $\color{blue}{\textsf{Schur}}$. Let $mn=x$, $np=y$, $pm=z$ to get $$*\iff x^3+y^3+z^3+3xyz\ge\sum\left(x^2y+xy^2\right).$$