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

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 ?

-
Please tell us where you are getting these unusual problems from! – Old John Jul 11 '12 at 13:56
I think, the inequality does not hold for $(a,b,c,d)=(1,1,1,2)$. – marlu Jul 11 '12 at 14:32
Hint: Try to prove that $$\frac{a^2+b^2+c^2}{a^5+b^5+c^5}\ge\frac{1}{abc}$$that is to say, $$a*b*c*(a^2+b^2+c^2)\ge（a^5+b^5+c^5）$$ – Pan Yan Jul 11 '12 at 14:59
@PanYan: Your hint is not true for $(a,b,c) = (1,1,2)$. – marlu Jul 11 '12 at 15:31
marlu you are right ,i edit it . – Frank Jul 11 '12 at 15:40

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.

-