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

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?

• What is $\sum\limits_{cyc}$; maybe cycle? Jan 13 at 22:33
• @TymaGaidash a cyclic sum. Basically $\sum_{cyc}a=a+b+c$ $(a\to b\to c\to a)$ depending on how many variables you have.
– PNT
Jan 14 at 11:39
• if you assume $x+y-z = 1,$ then how does that make the other denominators such as $y+z-x$ equal to $1$ ? My hunch is your initial C-S is too crude anyway. Jan 14 at 22:32
• even then, you are assuming $x+y-z = y+z-x = z+x-y = 1$ all simultaneously which is perhaps the most absurd thing I have seen in a while on this site. Jan 14 at 23:30
• You are assuming three conditions - you may only assume one. Do you understand why homogeneity allow us to make such assumptions in the first place? It's akin to normalizing the whole expression by the sum you are assuming to be one e.g,: $$x^2+y^2\geq 2xy \iff \dfrac{x^2}{(x+y)^2} + \dfrac{y^2}{(x+y)^2}\geq 2\dfrac{x}{x+y}\dfrac{y}{x+y}$$ and since $\dfrac{x}{x+y} + \dfrac{y}{x+y} = 1$, we can now proceed to assume $x+y = 1.$ Jan 15 at 15:56

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: $$\sum_{cyc}\frac{\sqrt{2xy\cos Z}}{x+y-z}\le\sum_{cyc}\left(\frac{x+y}{2}+2\cos Z-x-y+z\right)=2\sum_{cyc}\cos X\le3$$

• You can't just go and substitute $x^2+y^2−z^2/2xy=\cos Z$
– PNT
Jan 14 at 21:38
• I should have defined the triangle XYZ where $XY=z$, $YZ=x$ and $ZX=y$. Since $a+b>c$, $x^2+y^2>z^2$, the triangle is acute. Jan 14 at 22:05

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 $$\frac{1}{\sqrt a+\sqrt b+\sqrt c+\sqrt{b+c-a}}\geq\frac{1}{\sqrt a+\sqrt b+\sqrt c+\sqrt{a+c-b}}.$$

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 $$*\iff x^3+y^3+z^3+3xyz\ge\sum\left(x^2y+xy^2\right).$$

• You have to mention that $\sqrt{a}, \sqrt{b}$ and $\sqrt{c}$ are also sides of a triangle, otherwise yout substitution is not allowed .
– PNT
Jan 14 at 11:47
• @PNT I remember this substitution had a name, but I forgot it.
– user1034536
Jan 14 at 12:44