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Hi It's a generalization found on Aops starting from this question Prove: $\frac{\frac{1}{a}+b}{\sqrt{\frac{1}{a}+a}}+\frac{\frac{1}{b}+c}{\sqrt{\frac{1}{b}+b}}+\frac{\frac{1}{c}+a}{\sqrt{\frac{1}{c}+c}}\ge3\sqrt{2}$ (see comments) :

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

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

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

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

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

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

Edit :

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

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

Question :

How to achieve my work (if true) or show $(I)$?

Miss and Mister cassoulet char
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1 Answers1

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Lemma 1: For $a,b>0$, $$2\sqrt{\frac{1}{a}+b} \geqslant \sqrt2 \left(\frac{1}{\sqrt{a}}+\sqrt{b}\right)$$ Proof: $$\left(\frac{1}{\sqrt{a}}-\sqrt{b}\right)^2 \geqslant 0$$ $$\frac{1}a + b \geqslant 2\frac{\sqrt{b}}{\sqrt{a}}$$

$$\left(\frac{1}{\sqrt{a}}+\sqrt{b}\right)^2= \frac{1}a + b + 2\frac{\sqrt{b}}{\sqrt{a}} \leqslant 2\left(\frac{1}a + b\right) $$

and the result follows.

Lemma 2: For $a>0$, $$\sqrt{\frac{1}{a}+a} \leqslant \sqrt2 \left(\frac{1}{\sqrt{a}}+\sqrt{a}-1\right)$$ Proof: $$(\sqrt{a}-1)^4\geqslant 0$$ $$a^2-4\sqrt{a}a+6a-4\sqrt{a}+1\geqslant 0$$ $$2(a^2-2\sqrt{a}a+3a-2\sqrt{a}+1)\geqslant a^2+1$$ $$2(a-\sqrt{a}+1)^2\geqslant a^2+1$$ $$2\left(\sqrt{a}-1+\frac{1}{\sqrt{a}}\right)^2\geqslant a+\frac{1}{a}$$ and the result follows.

Lemma 3: For $a,b>0$, $$\frac{{\frac{1}{a}+b}}{\sqrt{\frac{1}{a}+a}} \geqslant \sqrt2 (\sqrt{b}-\sqrt{a}+1)$$ Proof: By AM-GM and then Lemma 1: $$\frac{\frac{1}{a}+b}{\sqrt{\frac{1}{a}+a}}+\sqrt{\frac{1}{a}+a} \geqslant 2 \sqrt{\frac{1}{a}+b} \geqslant \sqrt2 \left(\frac{1}{\sqrt{a}}+\sqrt{b}\right)$$

Combining with Lemma 2:

$$\frac{\frac{1}{a}+b}{\sqrt{\frac{1}{a}+a}}+ \sqrt2 \left(\frac{1}{\sqrt{a}}+\sqrt{a}-1\right) \geqslant \sqrt2 \left(\frac{1}{\sqrt{a}}+\sqrt{b}\right)$$ and the result follows.

Theorem: For $x_i>0$, $$\sum_{i=1}^{n}\frac{\frac{1}{x_i}+x_{i+1}}{\sqrt{\frac{1}{x_i}+x_i}}\geq n\sqrt{2}$$ Proof: Let $a=x_i$ and $b=x_{i+1}$ in Lemma 3 and sum.

Blitzer
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