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let $a,b,c,d>0$ show that $$\sqrt{\left(a+\sqrt{\dfrac{bcd}{a}}\right)\left(b+\sqrt{\dfrac{acd}{b}}\right)\left(c+\sqrt{\dfrac{abd}{c}}\right)\left(d+\sqrt{\dfrac{abc}{d}}\right)}+2\sqrt{abcd}\ge ab+bc+cd+da+ac+bd$$

My idea: since $$a+\sqrt{\dfrac{bcd}{a}}=a+\dfrac{\sqrt{abcd}}{a}\ge 2\sqrt{\sqrt{abcd}}$$

It is said this inequality can use AM-GM( Cauchy-Schwarz) inequality to solve it.But I can't.

Thank you for you help

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Why have you put Cauchy-Schwarz inside brackets?AM-GM is quite different from Cauchy-Schwarz. –  rah4927 Feb 6 at 6:18
    
As the inequality is homogenous, setting $abcd=1$ gives $$\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}+2\geq\sum_{sym}ab.$$ Cauchy-Schwarz becomes recognizable, but the difficult part seems to be the $+2$. Not sure how to go from here. I have a few ideas (If you want me to in can write them in an answer but I don't think they're very valuable.) –  barto Feb 7 at 22:22

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Solution: Use the following factorization $\prod_{cyc}(a^2+x^2) = \prod_{cyc}(x+ai)\prod_{cyc}(x-ai)$ to see that $$\prod_{cyc}(a^2+x^2) = (x^4+abcd - ab-ac - bc - cd - da-db )^2 + (x^3(a+b+c+d)-x(abc+bcd+cda+dab))^2$$, where x is arbitrary number. If you let $x = (abcd)^{\frac 14}$, and ignore the second term on the right hand side of the equality, then our desired inequality follows. Inequality holds if and only if $$(ab-cd)(ac-bd)(ad-bc) = 0$$

P.S: This inequality was proposed by me for the Mongolian National Math Olympiad for seniors in 2010. So I am fairly certain that it is an original inequality. In some sense, it is a stronger version of the following well known inequality. $$a^4+b^4+c^4+d^4+2abcd\geq a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2$$

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