Prove $\sum_{cyc}{\sqrt{\frac{x+1}{x^2+16x+1}}}\geqslant 1$ and $ \sum_{cyc}{\sqrt{\frac{x+1}{4x^2+10x+4}}}\leqslant 1$ for $x,y,z>0,xyz=1$

Question

Source: https://artofproblemsolving.com/community/c6t243f6h2745656_inequalities_with_3

Let $x,y,z>0,xyz=1$ then $$ \sum_{cyc}{\sqrt{\frac{x+1}{x^2+16x+1}}}\geqslant 1 \ \ \ \ (1) $$ and$$ \sum_{cyc}{\sqrt{\frac{x+1}{4x^2+10x+4}}}\leqslant 1\ \ \ (2) $$ For this type of inequalities, I usually use local inequalities$$ x,y,z>0,xyz=1,\sum_{cyc}{\frac{1}{x^2+x+1}}\geqslant 1, \sum_{cyc}{\frac{x^2}{x^2+x+1}}\geqslant 1, \sum_{cyc}{\frac{x+1}{x^2+x+1}}\leqslant 2 $$ For (1) it's smooth,because$$ \frac{x^4+1}{x^8+16 x^4+1}-\left(\frac{1}{x^2+x+1}\right)^2=\frac{(x-1)^2 x \left(2 x^4+7 x^3+14 x^2+7 x+2\right)}{\left(x^2+x+1\right)^2 \left(x^8+16 x^4+1\right)} \geqslant 0$$ Then it's easy to prove. But for (2) I get $$\frac{x^2+1}{4 x^4+10 x^2+4}-\left(\frac{x+1}{2 \left(x^2+x+1\right)}\right)^2=\frac{(x-1)^2 x^2}{4 \left(x^2+2\right) \left(x^2+x+1\right)^2 \left(2 x^2+1\right)} \geqslant 0$$ Then $$ \sum_{cyc}{\sqrt{\frac{x+1}{4x^2+10x+4}}}\geqslant \sum_{cyc}{\frac{1}{2}\frac{\sqrt{x}+1}{x+\sqrt{x}+1}}\leqslant 1 $$ Failed. How can I prove (2)? Any solution is welcome.

Some proofs are given in OP's link: https://artofproblemsolving.com/community/c6t243f6h2745656_inequalities_with_3 — River Li, Jan 12 '22 at 11:55

Miss and Mister cassoulet char · Answer 1 · 2022-02-12T12:22:04.097

0

Hint :

For $x\leq 0$ we have :

$$\frac{d}{dx}\left(\frac{d}{dx}f\left(e^{x}\right)\right)<0$$

Where :

$$f(x)=\sqrt{\frac{x+1}{4x^{2}+4+10x}}$$

Next we show that for $x>0$ and $n= 3$ a natural number :

$$f\left(x\right)+\left(n-1\right)f\left(\frac{1}{x^{\frac{1}{n-1}}}\right)-nf\left(1\right)\leq 0$$

In the case $n=3$ we can use the bound given in the link https://artofproblemsolving.com/community/c6t243f6h2745656_inequalities_with_3

We have for $x>0$:

$$\frac{1}{1+\sqrt{1+3x}}-\sqrt{\frac{x+1}{4x^{2}+4+10x}}\geq 0$$

To show this inequality we use the substitution $y^2=3x+1$ then square both side then clearing the denominator and use a factorization .

So we need to show :

$$\frac{1}{1+\sqrt{1+3x}}+2\frac{1}{1+\sqrt{1+3x^{-0.5}}}-3\frac{1}{1+\sqrt{1+3}}\leq 0$$

Wich is easy !

So now we can apply the LCF corollary (first reference) and conclude .

Last remark : We have also the case $n=4$ for $a_i\in(e^{-1},e^{1})$ such that $\prod_{i=1}^{4}a_i=1$.

Reference :

Cirtoaje and Baiesu: An extension of Jensen’s discrete inequality to half convex functions. Journal of Inequalities and Applications 2011 2011:101.

2)https://artofproblemsolving.com/community/c6t243f6h2745656_inequalities_with_3

Edit 11/02/2022 :

For $0<x\leq 1$ we have :

$$\frac{\left(x+2\right)\left(x+1\right)}{\frac{11}{2}x^{2}+4+\frac{17}{2}x}\geq f(x)$$

On the other hand we have $x\geq 1$ :

$$h\left(x\right)=\frac{2}{3}-\frac{1}{1.5x^{-0.5}+1.5}\geq f(x)$$

Edit 12/02/2022 :

We can do better on $x\in (0,1]$ we have :

$$r(x)=\frac{\frac{1}{2}x^{2}+\frac{7}{2}x+2}{4x^{2}+4+10x}\geq f(x)$$

Using the same approach as above we have on $(-\infty,0]$ :

$$\frac{d}{dx}\left(\frac{d}{dx}r\left(e^{x}\right)\right)<0$$

So we have for $0<a\leq 1$ and $0<b\leq 1$ and $c\geq 1$ :

$$f(a)+f(b)+f(c)\leq 2r(\sqrt{ab})+h(c)=2r(\sqrt{ab})+h(1/ab)\leq 1$$

Wich is smooth .

For the other case I use Buffalo's ways see Wolfram alpha

*I made a mistake I solve another problem ...

Here you can find the solution for the other case wich use Buffalo's way

See Wolfram alpha (2).

edited Feb 12 '22 at 12:22

answered Jan 02 '22 at 15:24

Miss and Mister cassoulet char

3,511

Would you please give some details, e.g. which theorem did you use? The paper is "Reciprocally convex functions" by MilanMerkle, Journal of Mathematical Analysis and Applications, Volume 293, Issue 1, 1 May 2004, Pages 210-218. – River Li Jan 09 '22 at 16:51
I think the paper does not give the result: $f\left(a\right)+f\left(b\right)+f\left(c\right)\leq 3f\left(\left(abc\right)^{\frac{1}{3}}\right)$. In Theorem 2.3, eq. (4), it only gives $f(a)+f(b)+f(c) \le 3f(\frac{3}{1/a + 1/b + 1/c})$ which is weaker than $f\left(a\right)+f\left(b\right)+f\left(c\right)\leq 3f\left(\left(abc\right)^{\frac{1}{3}}\right)$. Please let me know if I am wrong. – River Li Jan 12 '22 at 12:01
@RiverLi it's just the substitution $x=e^u$ see the page 215 of the same paper . – Miss and Mister cassoulet char Jan 12 '22 at 12:18
I don't understand you. Would you please write down the detailed proof? – River Li Jan 12 '22 at 12:50
In Theorem 3.1, you need the assumption $f(e^x)$ is concave on $(-\infty, \infty)$, or convex on $(-\infty, \infty)$. If you let $f(x) = \sqrt{\frac{\left(x+1\right)}{4x^2+10x+4}}$, then $f(e^x)$ is neither concave nor convex on $(-\infty, \infty)$. – River Li Jan 12 '22 at 12:55
1

@RiverLi thanks for help me now it's fixed with a partial proof ! – Miss and Mister cassoulet char Jan 12 '22 at 17:30
@RiverLi Can you show the inequality related to my edit ? – Miss and Mister cassoulet char Feb 08 '22 at 15:50
Is your edit a full proof for the original problem? If so, we are interested in it and you can write down the details. If no, we wait for a full proof. – River Li Feb 09 '22 at 01:07
@RiverLi see my edits ! – Miss and Mister cassoulet char Feb 09 '22 at 11:27
@RiverLi it's a variant does it matter ? – Miss and Mister cassoulet char Feb 09 '22 at 11:40
In your edit, if you have a different proof from the proofs in the link (but use the bound given in the link), you may write down details. For example, why "So we need to show"? – River Li Feb 09 '22 at 13:37
@I have not the time yet but no problem !:-) – Miss and Mister cassoulet char Feb 09 '22 at 13:59
@RiverLi To show the inequality in question I use the linear transformation $y=3x+1$ next square both side the rest is smooth.I guess you show the inequality like this so I'm thinking to another proof . For your second question see the first reference and LCF corollary . – Miss and Mister cassoulet char Feb 10 '22 at 10:09
1

You should write down the details (step-by-step). You mixed different content such as partial hint, edit, edit again. It is not nice. You may delete the partial hint and focus on your main proof. Or keep partial hint and post a new answer. – River Li Feb 10 '22 at 11:38
@RiverLi see my edits thanks for the advices . – Miss and Mister cassoulet char Feb 10 '22 at 14:20
It is hard for me to read your proof. First, you can focus on $n = 3$ and give remark (at the end or at the beginning) for general $n$. Second, how is the 1st paragrph "for $x \le 0$ we have $\frac{d}{dx}\left(\frac{d}{dx}f\left(e^{x}\right)\right)<0$" related to your proof? – River Li Feb 10 '22 at 15:48
By the way, I did not downvote your another answer. – River Li Feb 11 '22 at 11:25
@RiverLi see the condition related to the LCF corollary (first reference). – Miss and Mister cassoulet char Feb 11 '22 at 15:24
@RiverLi what do you think about my two last bound ? Thanks for your advices ! – Miss and Mister cassoulet char Feb 11 '22 at 16:19
If you provide a full proof, we are interested in it. – River Li Feb 12 '22 at 01:06
@RiverLi See my edits . Hope you can find something pleasant .:-) – Miss and Mister cassoulet char Feb 12 '22 at 11:32
It is not a full proof, right? Can you write down step-by-step proof for the OP? I don't understand your Edit 12/02/2022 at all. – River Li Feb 12 '22 at 12:07
@RiverLi The bounds are really simple to show and again it's a simple use of Jensen's inequality where I use the second derivative but I can add the details even if I have not the time today .See you tomorrow . – Miss and Mister cassoulet char Feb 12 '22 at 12:24
You should write down step-by-step solution (all details). You can do it in the future when you are not busy. – River Li Feb 12 '22 at 12:28

Prove $\sum_{cyc}{\sqrt{\frac{x+1}{x^2+16x+1}}}\geqslant 1$ and $ \sum_{cyc}{\sqrt{\frac{x+1}{4x^2+10x+4}}}\leqslant 1$ for $x,y,z>0,xyz=1$

1 Answers1

Linked