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Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: $$\sqrt{\frac{a}{4a+2b+3}}+\sqrt{\frac{b}{4b+2c+3}}+\sqrt{\frac{c}{4c+2a+3}}\leq1$$

The equality "occurs" also for $a\rightarrow+\infty$, $b\rightarrow+\infty$ and $a>>b$.

I tried AM-GM, C-S and more, but without any success.

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    This kind of cyclic but asymmetric inequalities is usually tackled through *mixing variables* or similar techniques. – 2017-01-27
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    @Jack D'Aurizio MV works for symmetric inequalities. I think it can not help here. – 2017-01-27
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    Are you sure about the "equality" for $a=b \to \infty$? Wouldn't the limit be $1/\sqrt 6 + 1/2 + 0 < 1$? – 2017-01-27
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    @Martin R I fixed my post. Thank you! – 2017-01-27
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    @MichaelRozenberg: it works for cyclic inequalities, too (gil.ro/downloadable/download/sample/sample_id/1/). It is practical because it can solve problems with multiple stationary points. – 2017-01-27
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    Equality will occur when $a=b=c=1$ – 2017-01-27
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    @Joseph Quarcoo Yes of course! – 2017-01-27
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    @Jack D'Aurizio I'll think about it. Thank you! – 2017-01-27
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    Hello I can transform your inequality in a cyclic inequality . I will post my answer tomorrow .It's a nice inequality thanks for that ! – 2017-06-03

1 Answers 1

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We begin with a theorem :

Theorem :

Let $a,b,c,d,e,f$ be positive real number , with $a\geq b \geq c$ , $d\geq e \geq f $ under the three following conditions :

$a\geq d$ , $ab\geq de$ , $abc\geq def$ so we have :

$$a+b+c\geq d+e+f$$

Here we suppose that we have :

$a\geq b \geq 1 \geq c $

So to get the majorization we prove this :

$\sqrt{\frac{a}{4a+2b+3}}\geq \sqrt{\frac{c}{4c+2a+3}}$

Wich is equivalent to :

$\frac{a}{4a+2b+3}\geq \frac{c}{4c+2a+3}$

Or :

$a(4c+2a+3)\geq c(4a+2b+3)$

Wich is obvious under the previous conditions.

With the same reasoning we can prove that we have :

$\sqrt{\frac{b}{4b+2c+3}}\geq \sqrt{\frac{c}{4c+2a+3}}$

Now we study the case :

$\sqrt{\frac{a}{4a+2b+3}}\geq\sqrt{\frac{b}{4b+2c+3}}\geq \sqrt{\frac{c}{4c+2a+3}}$ wich corresponding to $a\geq b \geq c$ in the initial theorem

And

$0.5-\frac{1}{8.2a}\geq 0.5-\frac{1}{8.2b}\geq \frac{1}{8.2a}+\frac{1}{8.2b} $ wich corresponding to $d\geq e \geq f $ in the initial theorem

It's clear that we have :

$\sqrt{\frac{a}{4a+2b+3}}\leq \sqrt{\frac{a}{4a+3}}\leq 0.5-\frac{1}{8.2a}$

And

$\sqrt{\frac{b}{4b+2c+3}}\leq \sqrt{\frac{b}{4b+3}}\leq 0.5-\frac{1}{8.2b}$

So we have :

$\sqrt{\frac{a}{4a+2b+3}}\sqrt{\frac{b}{4b+2c+3}}\leq (0.5-\frac{1}{8.2a})(0.5-\frac{1}{8.2b})$

And

$\sqrt{\frac{a}{4a+2b+3}}\sqrt{\frac{b}{4b+2c+3}}\sqrt{\frac{c}{4c+2a+3}}\leq (0.5-\frac{1}{8.2a})(0.5-\frac{1}{8.2b})(\frac{1}{8.2a}+\frac{1}{8.2b})$

Wich is true because we have with the condition $abc=1$

$$27\leq \prod_{cyc}\sqrt{4a+2b+3}$$

So now you just have to apply the theorem with this majorization .

The case $\sqrt{\frac{b}{4b+2c+3}}\geq \sqrt{\frac{a}{4a+2b+3}} \geq \sqrt{\frac{c}{4c+2a+3}}$ is the same.

And for the case $a\geq 1 \geq b \geq c$ you just have to make the following substitution $B=\frac{1}{b}$ to find the previous case $a\geq b \geq 1 \geq c $

Edit :

With the previous substitution the original inequality becomes with $a\geq b \geq 1 \geq c$ and $ac=b$:

$$\sqrt{\frac{ab}{4ab+2+3b}}+\sqrt{\frac{1}{4+2cb+3b}}+\sqrt{\frac{c}{4c+2a+3}}$$

We can briefly prove that we have :

$\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{1}{4+2cb+3b}}$

And

$\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{c}{4c+2a+3}}$

Now we study the case :

$\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{1}{4+2cb+3b}}\geq \sqrt{\frac{c}{4c+2a+3}}$ wich corresponding to $a\geq b \geq c$ in the initial theorem

And

$0.5-\frac{1}{11(ab)^2}\geq \frac{1}{3} \geq 1-\frac{1}{3}-(0.5-\frac{1}{11(ab)^2})$

wich corresponding to $d\geq e \geq f $ in the initial theorem

Now you just have to apply the theorem with this majorization .

The case $\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{c}{4c+2a+3}}\geq \sqrt{\frac{1}{4+2cb+3b}}$ works this the same majorization.

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    In the last your step the inequality changes after substitution $B=\frac{1}{b}$. – 2017-08-11
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    Yes I will edit I realize now my mistake – 2017-08-11
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    There are many mistakes. – 2017-08-11
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    Can you say me where please ? – 2017-08-11
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    I edit the proof . – 2017-08-11