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!

Solution 1:

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.

Solution 2:

Using Radon's inequality we need to show :

$$\frac{\left(\left(\frac{1}{a}+b\right)^{\frac{2}{3}}+\left(\frac{1}{b}+c\right)^{\frac{2}{3}}+\left(\frac{1}{c}+a\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}}{\sqrt{a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}}-3\sqrt{2}\geq 0$$

Can you end now ?

Follow up

Let $x_i>0$, $1\leq i\leq 3=n$, such that $x_1=x_{4}$ then we have :

$$\sum_{i=1}^{n}\frac{\frac{1}{x_i}+x_{i+1}}{\sqrt{\frac{1}{x_i}+x_i}}\geq 3\sqrt{2}\tag{I}$$

At first glance it seems that Am-Gm is too weak so as above I use Radon's inequality (see vivid edit) we need to show :

$$\frac{\left(\sum_{i=1}^{n}\left(\frac{1}{x_i}+x_{i+1}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}}{\sqrt{\sum_{i=1}^{n}\left(\frac{1}{x_i}+x_i\right)}}\geq 3\sqrt{2}$$

Then I used Minkowski's inequality with $p=\frac{2}{3}$ we need to show :

$$\frac{\left(\sum_{i=1}^{n}\left(\frac{1}{x_i}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}+\left(\sum_{i=1}^{n}\left(x_{i}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}}{\sqrt{\sum_{i=1}^{n}\left(\frac{1}{x_i}+x_i\right)}}\geq 3\sqrt{2}$$

Edit :

We can apply Jensen's inequality and we need to show :

$$\frac{2^{-\frac{1}{2}}\left(\sum_{i=1}^{n}\left(\frac{1}{x_{i}}\right)^{\frac{2}{3}}+\sum_{i=1}^{n}\left(x_{i}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}}{\sqrt{\sum_{i=1}^{n}\left(\frac{1}{x_{i}}+x_{i}\right)}}-3\sqrt{2}\geq 0$$

Idea to conclude for the case $n=3$:

the function $x>0$ and $a,b\geq 4$:

$$f(x)=\left(x^{\frac{2}{3}}+\frac{1}{x^{\frac{2}{3}}}+a\right)^{\frac{3}{2}}-\left(x+\frac{1}{x}+b\right)^{\frac{1}{2}}\cdot6$$

We have :

$$f(x)\geq f(1)$$

Using this lemma a judicious number of times the conclusion follow for the case $n=3$.