# How prove this inequality $\sqrt{\prod_{cyc}(a+\sqrt{\frac{bcd}{a}})}+2\sqrt{abcd} \ge ab+bc+cd+da+ac+bd$

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