Prove $\sqrt{\frac{x}{\left(x+y\right)\left(1+y\right)}}+\sqrt{\frac{y}{\left(x+y\right)\left(1+x\right)}}+\sqrt{\frac{xy}{(1+x)(1+y)}}>1$

Question

For $x, y, z > 0$, prove that $$\sqrt{\dfrac{x}{\left(x+y\right)\left(1+y\right)}}+\sqrt{\dfrac{y}{\left(x+y\right)\left(1+x\right)}}+\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}>1.$$

I tried to use the AM-GM inequality as in the previous question $$\dfrac{x}{\sqrt{x\left(x+y\right)\left(1+y\right)}}+\dfrac{y}{\sqrt{y\left(x+y\right)\left(1+x\right)}}+\dfrac{xy}{\sqrt{xy\left(1+x\right)\left(1+y\right)}} \geq \dfrac{2x}{2x+xy+y+y^2}+\dfrac{2y}{2y+xy+x+x^2}+\dfrac{2xy}{1+x+y+2xy}=2 \cdot \left( \dfrac{x}{2x+xy+y+y^2}+\dfrac{y}{2y+xy+x+x^2}+\dfrac{xy}{1+x+y+2xy}\right).$$

The question again is it enough to prove the inequality this way?

Answer 1

For positive variables let $1=z$.

Thus, by Holder $$LHS^2=\frac{\left(\sum\limits_{cyc}\sqrt{\frac{xy}{(x+z)(y+z)}}\right)^2\sum\limits_{cyc}x^2y^2(x+z)(y+z)}{\sum\limits_{cyc}x^2y^2(x+z)(y+z)}\geq\frac{(xy+xz+yz)^3}{\sum\limits_{cyc}x^2y^2(x+z)(y+z)}>1.$$

Answer 2

Another way.

For positive variables let $1=z$.

Thus, $$LHS^2=\left(\sum\limits_{cyc}\sqrt{\frac{xy}{(x+z)(y+z)}}\right)^2=\frac{\left(\sum\limits_{cyc}\sqrt{xy(x+y)}\right)^2}{\prod\limits_{cyc}(x+y)}=$$ $$=\frac{\sum\limits_{cyc}\left(x^2y+x^2z+2\sqrt{x^2yz(x+y)(x+z)}\right)}{\prod\limits_{cyc}(x+y)}>\frac{\sum\limits_{cyc}\left(x^2y+x^2z+\frac{2}{3}xyz\right)}{\prod\limits_{cyc}(x+y)}=1.$$

Answer 3

By AM-GM, we have \begin{align*} &\sqrt{\frac{x}{\left(x+y\right)\left(1+y\right)}}+\sqrt{\frac{y}{\left(x+y\right)\left(1+x\right)}}+\sqrt{\frac{xy}{\left(1+x\right)\left(1+y\right)}}\\[6pt] ={}& \frac{2x}{2\sqrt{(x + y) \cdot x(1 + y)}} + \frac{2y}{2\sqrt{(x + y)\cdot y(1 + x)}} + \frac{2xy}{2\sqrt{x(1 + y) \cdot y(1+x)}}\\[6pt] \ge{}& \frac{2x}{x+ y + x(1 + y)} + \frac{2y}{x + y + y(1 + x)} + \frac{2xy}{x(1+y) + y(1+x)}\\[6pt] ={}& \frac{2x}{2x + y + xy} + \frac{2y}{x + 2y + xy} + \frac{2xy}{x + y + 2xy}\\[6pt] >{}& \frac{2x}{2x + 2y + 2xy} + \frac{2y}{2x + 2y + 2xy} + \frac{2xy}{2x + 2y + 2xy} \\[6pt] ={}& 1. \end{align*}

We are done.