3

Let $x_{1},x_{2},\cdots,x_{n}>0$,show that $$n\left(\dfrac{3-2\sqrt{2}}{2\sqrt{2}}\right)\min_{1\le i\le n}\sqrt{x_{i}} +\left(1-\dfrac{1}{2\sqrt{2}}\right)\sum_{i=1}^{n}\sqrt{x_{i}}\le\sum_{i=1}^{n}\dfrac{x_{i}}{\sqrt{x_{i}+x_{i+1}}},x_{n+1}=x_{1}$$

It is said this inequality is proved by A.Kovacec. So I was looking for it ,and maybe the proof is in this paper:Two Contributions to Inequalities by Alexander Kovaćec. But unfortunately I dont't have permission to open this article, so I don't know if this question comes from this article at all. What I am more concerned about is how to prove the inequality.

math110
  • 93,304
  • 2
    In RHS, $x_{i+1}$ is undefined when $i=n$, how do you deal with it? – i9Fn Jun 21 '17 at 20:30
  • Hello in fact it's a direct application of weighted Karamata inequality . –  Jun 23 '17 at 09:46
  • I don't have the time to create an answer but look at the theorem 3 from this link .Take the following fonction : $$\phi(x)=\frac{x^2}{\sqrt{1+x^2}}$$ this function is convex and increasing .The $y_i$ are define like this : $y_i=\frac{x_i}{x_{i+1}}$ and the $x_i$ like this : $x_i=1$ the $p_i=x_{i+1}$ . For the majorization there is no problem if you reorder the differents variables . If you have questions tell me . –  Jun 23 '17 at 11:46

1 Answers1

0

I believe the following would help.

After replacing $x_{i}\rightarrow x_{i}^2$ we need to prove that $$\sum_{i=1}^n\frac{x_i^2}{\sqrt{x_i^2+x_{i+1}^2}}\geq\frac{(2\sqrt2-1)\sum\limits_{i=1}^nx_i}{2\sqrt2}+\frac{(3-2\sqrt2)n\min\limits_i{x_i}}{2\sqrt2}$$ or

$$\sum_{i=1}^n\left(\frac{x_i^2}{\sqrt{x_i^2+x_{i+1}^2}}-\frac{x_i}{\sqrt2}\right)\geq\frac{(2\sqrt2-3)\sum\limits_{i=1}^nx_i}{2\sqrt2}+\frac{(3-2\sqrt2)n\min\limits_i{x_i}}{2\sqrt2}$$ or $$\sum_{i=1}^n\frac{x_i\left(\sqrt2x_i-\sqrt{x_i^2+x_{i+1}^2}\right)}{\sqrt{x_i^2+x_{i+1}^2}}+\frac{3-2\sqrt2}{2}\left(\sum\limits_{i=1}^nx_i-n\min\limits_i{x_i}\right)\geq0$$ or $$\sum_{i=1}^n\frac{x_i(x_i^2-x_{i+1}^2)}{\sqrt{x_i^2+x_{i+1}^2}\left(\sqrt2x_i+\sqrt{x_i^2+x_{i+1}^2}\right)}+\frac{3-2\sqrt2}{2}\left(\sum\limits_{i=1}^nx_i-n\min\limits_i{x_i}\right)\geq0$$ or $$\sum_{i=1}^n\left(\frac{x_i(x_i^2-x_{i+1}^2)}{\sqrt{x_i^2+x_{i+1}^2}\left(\sqrt2x_i+\sqrt{x_i^2+x_{i+1}^2}\right)}-\frac{x_i-x_{i+1}}{2}\right)\geq\frac{2\sqrt2-3}{2}\left(\sum\limits_{i=1}^nx_i-n\min\limits_i{x_i}\right)$$ or

$$\sum_{i=1}^n\frac{(x_i-x_{i+1})\left(x_i^2-x_{i+1}^2+2x_ix_{i+1}-x_i\sqrt{2(x_i^2+x_{i+1}^2)}\right)}{\sqrt{x_i^2+x_{i+1}^2}\left(\sqrt2x_i+\sqrt{x_i^2+x_{i+1}^2}\right)}+$$ $$+(3-2\sqrt2)\left(\sum\limits_{i=1}^nx_i-n\min\limits_i{x_i}\right)\geq0$$ or $$\sum_{i=1}^n\frac{(x_i-x_{i+1})^2\left(x_i+x_{i+1}-\frac{\sqrt2x_i(x_i+x_{i+1})}{\sqrt2x_{i+1}+\sqrt{x_i^2+x_{i+1}^2}}\right)}{\sqrt{x_i^2+x_{i+1}^2}\left(\sqrt2x_i+\sqrt{x_i^2+x_{i+1}^2}\right)}+(3-2\sqrt2)\left(\sum\limits_{i=1}^nx_i-n\min\limits_i{x_i}\right)\geq0$$ or $$\sum_{i=1}^n\tfrac{(x_i-x_{i+1})^2(x_i+x_{i+1})\left(1-\frac{\sqrt2x_i}{\sqrt2x_{i+1}+\sqrt{x_i^2+x_{i+1}^2}}\right)}{\sqrt{x_i^2+x_{i+1}^2}\left(\sqrt2x_i+\sqrt{x_i^2+x_{i+1}^2}\right)}+(3-2\sqrt2)\left(\sum\limits_{i=1}^nx_i-n\min\limits_i{x_i}\right)\geq0.$$ Maybe from here you'll end it.

By the way, I have a proof for the following. $$\sum_{i=1}^n\tfrac{(x_i-x_{i+1})^2(x_i+x_{i+1})\left(1-\frac{\sqrt2x_i}{\sqrt2x_{i+1}+\sqrt{x_i^2+x_{i+1}^2}}\right)}{\sqrt{x_i^2+x_{i+1}^2}\left(\sqrt2x_i+\sqrt{x_i^2+x_{i+1}^2}\right)}+0.42\left(\sum\limits_{i=1}^nx_i-n\min\limits_i{x_i}\right)\geq0.$$

For $n=3$ the following stronger inequality is true.

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt2}$$

Michael Rozenberg
  • 194,933