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.

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
$$
\frac1{1-\sqrt{xy}}+\frac1{1-\sqrt{yz}}+\frac1{1-\sqrt{zx}}\le\frac1{1-x}+\frac1{1-y}+\frac1{1-z}\tag4
$$
