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Let $a,b,c\ge0$ such that: $(a+b)(b+c)(c+a)=1$.

Find the minimum value of:

$$P=\frac{\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ac(c+a)}}{\sqrt{ab+bc+ca}}$$

I've tried many things but all failed. Please help. Thank you.

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  • $\begingroup$ Since, by Cauchy-Schwarz, $a=b=c$ gives the maximum of this expression, I'd suggest looking for extreme cases like $a=0$, $b=c=\frac1{\sqrt[3]2}$. My guess would be $\sqrt[3]2$. $\endgroup$ Commented Jan 24, 2014 at 14:01

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If $c=0$ and $a=b=\frac{1}{\sqrt[3]2}$ then $P=\sqrt[3]2$.

We'll prove that it's a minimal value of $P$.

Indeed, we need to prove that $$\frac{\sum\limits_{cyc}\sqrt{ab(a+b)}}{\sqrt{ab+ac+bc}}\geq\sqrt[6]{4(a+b)(a+c)(b+c)}.$$ Now, by AM-GM $$\sum\limits_{cyc}\sqrt{ab(a+b)}=\sqrt{\sum_{cyc}\left(a^2b+a^2c+2\sqrt{a^2bc(a+b)(a+c)}\right)}\geq$$ $$\geq\sqrt{\sum_{cyc}\left(a^2b+a^2c+4abc\right)}.$$ Thus, it remains to prove that $$\frac{\sum\limits_{cyc}(a^2b+a^2c+4abc)}{ab+ac+bc}\geq\sqrt[3]{4(a+b)(a+c)(b+c)}$$ or $$a+b+c+\frac{9abc}{ab+ac+bc}\geq\sqrt[3]{4(a+b)(a+c)(b+c)}.$$ But by Schur $$a+b+c+\frac{9abc}{ab+ac+bc}\geq\sqrt[3]{(a+b+c)^3+\frac{27abc(a+b+c)^2}{ab+ac+bc}}\geq$$ $$\geq\sqrt[3]{(a+b+c)^3+81abc}=\sqrt[3]{\sum_{cyc}(a^3-a^2b-a^2c+abc)+\sum_{cyc}(4a^2b+4a^2c+28abc)}\geq$$ $$\geq\sqrt[3]{\sum_{cyc}\left(4a^2b+4a^2c+\frac{8}{3}abc\right)}=\sqrt[3]{4(a+b)(a+c)(b+c)}$$ and we are done!

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