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Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: $$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\geq\frac{9}{\sqrt{a+b+c+15}}$$

It seems nice enough.

I proved this inequality by Holder, but it quits very ugly.

Maybe there is something nice? Thank you!

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  • $\begingroup$ Can this inequality be useful?$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$$....See if this can be used anyway.... $\endgroup$ – tatan Sep 30 '16 at 6:41
  • $\begingroup$ No! Nesbitt is very weak here. $\endgroup$ – Michael Rozenberg Sep 30 '16 at 6:51
  • $\begingroup$ Isn't $a=b=c=1$ the solution that maximizes the RHS and minimizes the LHS? $\endgroup$ – Alex Silva Sep 30 '16 at 11:48
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    $\begingroup$ @AlexSilva No, that is not the case. The maximum of the RHS is $ 3 / \sqrt{2} $, which does indeed occur when $a = b = c = 1$. However, the LHS can be smaller. Take for instance $a = b = x$ and $c = 1/x^2$. Then the LHS is $2\sqrt{x^3 / (x^3 + 1)} + \sqrt{1/(2x^3)}$, which tends to $2$ as $x \to \infty$. And $2 < 3/\sqrt{2}$. $\endgroup$ – Dylan Oct 2 '16 at 0:35
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    $\begingroup$ Concerning the LHS, 2 is indeed the smallest value one can get. There is a general proof that, for $0 < k < 1$, one has $(\frac{a}{b+c})^k+(\frac{b}{c+a})^k+(\frac{c}{a+b})^k\geq$ min $(2 ; \frac{3}{2^k})$ given in Pham Kim Hung's book "Secrets in Inequalities (volume 2)", pp. 284 ff. $\endgroup$ – Andreas Oct 2 '16 at 15:34
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Put the following substitution :

$\frac{a}{b+c}=x$$\quad$$\frac{b}{a+c}=y$$\quad$$\frac{c}{b+a}=z$

Remark that :

$$a+b+c=((\frac{1}{x}+1)(\frac{1}{y}+1)(\frac{1}{z}+1))^{\frac{1}{3}}$$

And :

$$abc=1 \iff -2=-\frac{1}{xyz}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$

So we get the inequality related to this Post that you have proved by your own with brio .

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$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\geq\frac{9}{\sqrt{a+b+c+15}}\iff\sqrt{a+b+c+15}\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\right)\ge9$$
Making $c=\frac{1 } {ab}$ the expression becomes $$f(a,b)= \left(\sqrt{\frac{a^2b}{ab^2+1}}+\sqrt{\frac{ab^2}{a^2b+1}}+\sqrt{\frac{1}{ab(a+b)}}\right)\sqrt{\frac{ab(a+b+15)+1}{ab}}\ge9$$ for all positive $a,b$.

It follows $$\sqrt{\frac{a^3b+a^2b^2+15a^2b+a}{ab^2+1}}+\sqrt{\frac{ab^3+a^2b^2+15ab^2+b}{a^2b+1}}+\frac{1}{ab}\sqrt{\frac{ab(a+b+15)+1}{a+b}} \ge9$$ It is clear $f(x,y)$ has no maximum and, in order to prove the inequality, we want to get the minimum of $f(x,y)$.

This minimum can be calculated as usually for two variables ($f_x(x,y)=0$ and $f_y(x,y)=0$, etc).

We calculate as follows: since $f(a,b)=f(b,a)$ the minimum of $f(a,b)$ is equal to the minimum of $f(a,a)$ where $a\gt 0$. Hence we calculate the minimum of the function of one variable $$f(x,x)=2\sqrt{\frac{2x^4+15x^3+x}{x^3+1}}+\frac{1}{x^2}\sqrt{\frac{2x^3+15x^2+1}{2x}}$$

The calculation is straightforward although somewhat tedious giving the minimum $9$ for $x=1$. For further explanation, see figure below wherein the calculation (Wolfram) and the graph of the function (Desmos) confirm the result. Thus this minimum is attained with $a = b = c = 1$ and the proposed inequality is valid for all positive with $abc=1$.

enter image description here

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  • $\begingroup$ You are welcome to show us your calculations with derivatives. $\endgroup$ – Michael Rozenberg Oct 5 '16 at 17:07

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