Inequality $\frac{a^2+b^2+c^2}{a^5+b^5+c^5}+\cdots+\frac{d^2+a^2+b^2}{d ^5+a^5+b^5}\le\frac{a+b+c+d}{abcd}$

Question

Let:$a,b,c,d>0$ be real numbers ,how to prove that :

$$\frac{a^2+b^2+c^2}{a^5+b^5+c^5}+\frac{b^2+c^2+d^2}{b^5+c^5+d^5}+\frac{c^2+d^2+a^2}{c^5+d^5+a^5}+\frac{d^2+a^2+b^2}{d ^5+a^5+b^5}\le\frac{a+b+c+d}{abcd}$$.

Edit : I think I proved it. From Cauchy inequality we have $$ \left(\sum\limits_{cyc}x^5\right)\left(\sum\limits_{cyc}x\right)\geq \left(\sum\limits_{cyc}x^3\right)^2 $$ From Chebyshev inequality it follows $$ \left(\sum\limits_{cyc}x\right)\left(\sum\limits_{cyc}x^2\right)\leq 3\left(\sum\limits_{cyc}x^3\right)^2 $$ hence $$ \frac{\left(\sum\limits_{cyc}x^5\right)}{\left(\sum\limits_{cyc}x^2\right)}= \frac{\left(\sum\limits_{cyc}x^5\right)\left(\sum\limits_{cyc}x\right)}{\left(\sum\limits_{cyc}x\right)\left(\sum\limits_{cyc}x^2\right)}\leq \frac{3}{\left(\sum\limits_{cyc}x^3\right)}\leq \frac{1}{xyz} $$ In the last step I used AM-GM inequality. The rest is clear.

Is there a different way to prove it ?

Answer 1

This question can be solved only using the AM-GM inequality. Building on an idea that Pan Yang suggests in his comment, it suffices to show the following.

$$\frac{a^2+b^2+c^2}{a^5+b^5+c^5}\le\frac{d}{abcd}=\frac{1}{abc}$$

This is equivalent to showing that

$$a^3 bc+b^3ca+c^3ab\le a^5+b^5+c^5 $$

However, this is true by taking a weighted AM-GM as such (I'm writing it in full):

$$\frac{1}{5}a^5+\frac{1}{5}a^5+\frac{1}{5}a^5+\frac{1}{5}b^5+\frac{1}{5}c^5\ge(a^{15}b^{5}c^{5})^{\frac{1}{5}}=a^3bc$$

and similarly for the terms $b^3ca, c^3ab$. This completes the proof.

Answer 2

Also, by Muirhead(or Chebyshov) and AM-GM we obtain: $$\sum_{cyc}\frac{a^2+b^2+c^2}{a^5+b^5+c^5}\leq\sum_{cyc}\frac{3}{a^3+b^3+c^3}\leq\sum_{cyc}\frac{1}{abc}=\frac{a+b+c+d}{abcd}.$$