# Wild inequality

I'm trying to solve this elementary inequality, but so far no clue. Can anybody solve it? I tried am/gm inequality on both side, but yield the same result hence no inequality. Also I tried other tricks, but no answer. Any proofs or hint would be appreciated.This is the problem:

assume $a,b,c,d$ are real positive numbers i.e. $a,b,c,d, \in \mathbb{R}^+$. prove that:

$$a^4b + b^4c + c^4d+ d^4a \ge abcd(a+b+c+d).$$

• – Dr. Sonnhard Graubner Aug 5 '17 at 17:29
• the proof is not so simple – Dr. Sonnhard Graubner Aug 5 '17 at 17:29
• does this help you? – Dr. Sonnhard Graubner Aug 5 '17 at 17:32
• xD .thanks man! that was the proof using weighted am/gm. I also saw the other way of proof. thank you very much. – K.K.McDonald Aug 5 '17 at 17:34
• i don't see this way yet – Dr. Sonnhard Graubner Aug 5 '17 at 17:36

We need to prove that $$\sum_{cyc}\frac{a^3}{cd}\geq a+b+c+d.$$ Now, by C-S $$\sum_{cyc}\frac{a^3}{cd}=\sum_{cyc}\frac{a^4}{acd}\geq\frac{(a^2+b^2+c^2+d^2)^2}{abc+abd+acd+bcd}.$$ Thus, it remains to prove that $$\frac{(a^2+b^2+c^2+d^2)^2}{abc+abd+acd+bcd}\geq a+b+c+d$$ or $$\sum_{cyc}a^4+\frac{1}{2}\sum_{sym}a^2b^2\geq4abcd+\frac{1}{2}\sum_{sym}a^2bc,$$ which is true by Muirhead because $(4,0,0,0)\succ(1,1,1,1)$ and $(2,2,0,0)\succ(2,1,1,0)$.