Prove: $\frac{\frac{1}{a}+b}{\sqrt{\frac{1}{a}+a}}+\frac{\frac{1}{b}+c}{\sqrt{\frac{1}{b}+b}}+\frac{\frac{1}{c}+a}{\sqrt{\frac{1}{c}+c}}\ge3\sqrt{2}$ 
Let $a,b,c>0$. Prove that: $$\frac{\frac{1}{a}+b}{\sqrt{\frac{1}{a}+a}}+\frac{\frac{1}{b}+c}{\sqrt{\frac{1}{b}+b}}+\frac{\frac{1}{c}+a}{\sqrt{\frac{1}{c}+c}}\ge3\sqrt{2}$$

Anyone help me a hint to solve above problem?
I tried by AM-GM without success: $$LHS\ge3\sqrt[3]{\frac{\left(\frac{1}{a}+b\right)\left(\frac{1}{b}+c\right)\left(\frac{1}{c}+a\right)}{\sqrt{\left(\frac{1}{a}+a\right)\left(\frac{1}{b}+b\right)\left(\frac{1}{c}+c\right)}}}$$
The rest is proving: $$\left(\frac{1}{a}+b\right)\left(\frac{1}{b}+c\right)\left(\frac{1}{c}+a\right)\ge2\sqrt{2\left(\frac{1}{a}+a\right)\left(\frac{1}{b}+b\right)\left(\frac{1}{c}+c\right)}$$ which is not true by a=60,b=0.02 and c=0.9
Thank you for your help!

Lemma: From this problem that the OP posed (with terms rearranged),for $ a_i > 0$,
$$ \frac{ \frac{1}{ a_i } + a_{i+1} } { \sqrt{ \frac{1}{ a_i } + a_{i} } } \geq \sqrt{2} ( \sqrt{a_{i+1}} - \sqrt{a_i} + 1 ). $$
Corollary: This inequality follows by summing up over $a_i$, and the RHS simplifies to $ 3 \sqrt{2} $.

Notes

*

*Furthermore, we get the obvious generalization to $n$ variables, $ \sum \frac{ \frac{1}{ a_i } + a_{i+1} } { \sqrt{ \frac{1}{ a_i } + a_{i} } } \geq n \sqrt{2}$.

*Guessing this lemma isn't obvious to me. Proving it also requires some manipulation.

