How to Prove the IMO Inequality Challenge for Positive Reals?

[anemone]

For positive reals $a,\,b,\,c$, prove that \(\displaystyle \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\gt 2\).

 

[Euge]

Here is my solution.

Spoiler

Let $S = a + b + c$, $x = a/S$, $y = b/S$, and $z = c/S$, so that the expression may be written

$$\sqrt{\frac{x}{1-x}} + \sqrt{\frac{y}{1-y}} + \sqrt{\frac{z}{1-z}}$$

Let $u$, $v$, and $w$ be positive numbers such that $u^2 = x/(1 - x)$, $v^2 = y/(1 - y)$, and $w^2 = z/(1 - z)$. Then $x = u^2/(u^2 + 1)$, $y = v^2/(v^2 + 1)$, and $z = w^2/(w^2 + 1)$. Note that neither $u$, $v$, nor $w$ is equal to $1$, otherwise $x = y = z = 1/2$, implying that $x + y + z = 3/2$, even though $x + y + z = 1$. Therefore, $v^2 + 1 > 2v$, $v^2 + 1 > 2v$, and $w^2 + 1 > 2w$, forcing $x < \frac{u}{2}$, $y < \frac{v}{2}$, and $z < \frac{w}{2}$. Hence,

$$\sqrt{\frac{x}{1 - x}} + \sqrt{\frac{y}{1 - y}} + \sqrt{\frac{z}{1 - z}} = u + v + w > 2x + 2y + 2z = 2(x + y + z) = 2$$

 

[anemone]

Here is my solution.

Spoiler

Let $S = a + b + c$, $x = a/S$, $y = b/S$, and $z = c/S$, so that the expression may be written

$$\sqrt{\frac{x}{1-x}} + \sqrt{\frac{y}{1-y}} + \sqrt{\frac{z}{1-z}}$$

Let $u$, $v$, and $w$ be positive numbers such that $u^2 = x/(1 - x)$, $v^2 = y/(1 - y)$, and $w^2 = z/(1 - z)$. Then $x = u^2/(u^2 + 1)$, $y = v^2/(v^2 + 1)$, and $z = w^2/(w^2 + 1)$. Note that neither $u$, $v$, nor $w$ is equal to $1$, otherwise $x = y = z = 1/2$, implying that $x + y + z = 3/2$, even though $x + y + z = 1$. Therefore, $v^2 + 1 > 2v$, $v^2 + 1 > 2v$, and $w^2 + 1 > 2w$, forcing $x < \frac{u}{2}$, $y < \frac{v}{2}$, and $z < \frac{w}{2}$. Hence,

$$\sqrt{\frac{x}{1 - x}} + \sqrt{\frac{y}{1 - y}} + \sqrt{\frac{z}{1 - z}} = u + v + w > 2x + 2y + 2z = 2(x + y + z) = 2$$

Very well done, Euge! And thanks for participating!(Cool)

Here is my solution:

Spoiler

First we multiply the top and bottom of each fraction by \(\displaystyle \sqrt{a},\,\sqrt{b},\,\sqrt{c}\) respectively, and simplify along the way to get:

\(\displaystyle \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\)

\(\displaystyle =\frac{\sqrt{a}}{\sqrt{b+c}}+\frac{\sqrt{b}}{\sqrt{c+a}}+\frac{\sqrt{c}}{\sqrt{a+b}}\)

\(\displaystyle =\frac{\sqrt{a}\sqrt{a}}{\sqrt{a}\sqrt{b+c}}+\frac{\sqrt{b}\sqrt{b}}{\sqrt{b}\sqrt{c+a}}+\frac{\sqrt{c}\sqrt{c}}{\sqrt{c}\sqrt{a+b}}\)

\(\displaystyle =\frac{a}{\sqrt{a}\sqrt{b+c}}+\frac{b}{\sqrt{b}\sqrt{c+a}}+\frac{c}{\sqrt{c}\sqrt{a+b}}\)

\(\displaystyle \gt \frac{a}{\frac{a+b+c}{2}}+\frac{b}{\frac{a+b+c}{2}}+\frac{c}{\frac{a+b+c}{2}}\) (from the AM-GM inequality)

\(\displaystyle \gt 2\left(\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}\right)\)

\(\displaystyle = 2\left(\frac{a+b+c}{a+b+c}\right)\)

\(\displaystyle = 2 \)(Q.E.D.)