Prove that $\frac{1}{1 - \sqrt{ab}} + \frac{1}{1 - \sqrt{bc}} + \frac{1}{1 - \sqrt{ca}} \leq \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$

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.

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

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