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Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that: $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq\sqrt{\frac{24(a^2+b^2+c^2+d^2)}{ab+ac+bc+ad+bd+cd}}$$

This inequality is similar to the following inequality of three variables.

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq3\sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}},$$ which after squaring of the both sides is equivalent to $$\sum_{cyc}\left(\frac{a^3}{b}+\frac{a^3c}{b^2}+2ab+\frac{2a^2b}{c}+\frac{a^2c}{b}-7a^2\right)\geq0,$$ which is just AM-GM.

This method does not help for the starting inequality.

A tried also another ways, but without any success.

For example, by Holder $$\left(\sum_{cyc}\frac{a}{b}\right)^2\sum_{cyc}a^4b^2\geq\left(\sum_{cyc}a^2\right)^3.$$ Hence, it remains to prove that $$(a^2+b^2+c^2+d^2)^2(ab+ac+bc+ad+bd+cd)\geq24(a^4b^2+b^4c^2+c^4d^2+d^4a^2),$$ which is wrong for $c=d\rightarrow0^+$.

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    after squaring it can be proven by the substitution $$b=a+u,c=a+u+v,d=a+u+v+w$$ – 2017-04-08
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    Sonnhard, it's hard for me. – 2017-04-08
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    yes for me too but it can be solved i'm here in Vienna for a short visit – 2017-04-08
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    @Dr.SonnhardGraubner Would you mind to work out your first comment and give an answer to the OP? – 2017-12-08

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