Prove that $\frac{1}{1 - \sqrt{ab}} + \frac{1}{1 - \sqrt{bc}} + \frac{1}{1 - \sqrt{ca}} \leq \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$ Given that $0 < a , b , c < 1$.
Prove that $\frac{1}{1 -  \sqrt{ab}} + \frac{1}{1 - \sqrt{bc}} + \frac{1}{1 - \sqrt{ca}} \leq \frac{1}{1 -  a} + \frac{1}{1 - b} + \frac{1}{1 - c}$.
I tried using modified C.S. and brute-force. But , it demands a lot of calculation. So , I want some better solution than this. Thank you.
 A: The RHS is
$$\sum_{n=0}^\infty(a^n+b^n+c^n).$$
The LHS is
$$\sum_{n=0}^\infty(\sqrt{a^nb^n}+\sqrt{a^nc^n}+\sqrt{b^nc^n}).$$
For each $n$,
$$a^n+b^n+c^n=\frac{a^n+b^n}2+\frac{a^n+c^n}2+\frac{b^n+c^n}2
\ge\sqrt{a^nb^n}+\sqrt{a^nc^n}+\sqrt{b^nc^n}$$
by AM/GM
A: This is very similar to N.Quy's approach, but bringing out the monotonicity and convexity of $\frac1{1-x}$ made this inequality clearer for me.
Since $\frac1{1-x}$ is monotonically increasing on $[0,1)$, the AM-GM says that
$$
\frac1{1-\sqrt{xy}}\le\frac1{1-\frac{x+y}2}\tag1
$$
Since $\frac1{1-x}$ is convex on $[0,1)$, we have
$$
\frac1{1-\frac{x+y}2}\le\frac12\left(\frac1{1-x}+\frac1{1-y}\right)\tag2
$$
Therefore,
$$
\frac1{1-\sqrt{xy}}\le\frac12\left(\frac1{1-x}+\frac1{1-y}\right)\tag3
$$
Adding $(3)$ for all $3$ pairs gives
$$
\frac1{1-\sqrt{xy}}+\frac1{1-\sqrt{yz}}+\frac1{1-\sqrt{zx}}\le\frac1{1-x}+\frac1{1-y}+\frac1{1-z}\tag4
$$
A: We have $\sqrt{ab}\leq \frac{a+b}{2}$ so $1-\sqrt{ab}\geq 1-\frac{a+b}{2}>0$, which implies
$$\dfrac{1}{1-\sqrt{ab}}\leq \dfrac{2}{2-a-b}$$
And using the simple inequality $\frac{4}{x+y}\leq \frac{1}{x}+\frac{1}{y}$ for all $x,y>0$ you get
$$\dfrac{1}{1-\sqrt{ab}}\leq \dfrac{2}{2-a-b}\leq \dfrac{1}{2(1-a)}+\dfrac{1}{2(1-b)}$$
Summing up gives the result.
