Prove that for positive numbers x and y the inequality is true.

Question

Prove that for positive numbers x and y the inequality $$ \sqrt{\dfrac{1}{x+y}}+\sqrt{\dfrac{x}{1+y}}+\sqrt{\dfrac{y}{1+x}}>2$$ is true.

I tried to use the inequality $$\dfrac{1}{\sqrt{ab}} \geq \dfrac{2}{a+b}.$$

For example, the first fraction turns to $$\sqrt{\dfrac{1}{(x+y)1}} \geq \dfrac{2}{x+y+1}.$$

The second fraction turns to $$\sqrt{\dfrac{x}{(1+y)1}} \geq \dfrac{2x}{y+2}.$$

The third fraction turns to $$\sqrt{\dfrac{y}{(1+x)1}} \geq \dfrac{2y}{x+2}.$$

So we have $$ \sqrt{\dfrac{1}{x+y}}+\sqrt{\dfrac{x}{1+y}}+\sqrt{\dfrac{y}{1+x}} \geq \dfrac{2}{x+y+1}+\dfrac{2x}{2+y}+\dfrac{2y}{2+x} = 2 \left( \dfrac{1}{x+y+1}+\dfrac{x}{2+y}+\dfrac{y}{2+x}\right)>2.$$ Is it enough to prove that multiplying by 2, we get an expression greater than two?

Any hint would help a lot, Thanks!

Answer 1

Let $1=z$.

Thus, by Holder and Schur we obtain: $$LHS=\sum_{cyc}\sqrt{\frac{x}{y+z}}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{\frac{x}{y+z}}\right)^2\sum\limits_{cyc}x^2(y+z)}{\sum\limits_{cyc}x^2(y+z)}}\geq\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}x^2(y+z)}}>2.$$

Answer 2

Another way. By AM-GM $$\sqrt{\dfrac{1}{x+y}}+\sqrt{\dfrac{x}{1+y}}+\sqrt{\dfrac{y}{1+x}}=\dfrac{1}{\sqrt{1(x+y)}}+\dfrac{x}{\sqrt{x(1+y)}}+\dfrac{y}{\sqrt{y(1+x)}}\geq$$ $$\geq\frac{2}{1+x+y}+\frac{2x}{x+1+y}+\frac{2y}{y+1+x}=2.$$ The equality occurs for $1=x+y$, $x=1+y$ and $y=1+x,$ which says that the equality does not occur.