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.


1 Answer 1


Hint :

For $x\leq 0$ we have :


Where :


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 :

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


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 $(-\infty,0]$ :


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).

  • $\begingroup$ 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. $\endgroup$
    – River Li
    Jan 9, 2022 at 16:51
  • $\begingroup$ 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. $\endgroup$
    – River Li
    Jan 12, 2022 at 12:01
  • $\begingroup$ @RiverLi it's just the substitution $x=e^u$ see the page 215 of the same paper . $\endgroup$ Jan 12, 2022 at 12:18
  • 1
    $\begingroup$ @RiverLi thanks for help me now it's fixed with a partial proof ! $\endgroup$ Jan 12, 2022 at 17:30
  • 1
    $\begingroup$ 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. $\endgroup$
    – River Li
    Feb 10, 2022 at 11:38

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