Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$ Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$
My try
My book gives as a hint   to move everything to the left hand side of the inequality and then factor and see what I get in the long factorization process and to lookout  for squares. 
So that's what I have tried:
\begin{array}
((a^7+b^7)(a^2+b^2) &\ge (a^5+b^5)(a^4+b^4) \\\\
(a^7+b^7)(a^2+b^2)-(a^5+b^5)(a^4+b^4) &\ge 0 \\\\
(a+b)(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)(a^2+b^2)-(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)(a^4+b^4) &\ge 0 \\\\
(a+b)\left[(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)((a+b)^2-2ab)-(a^4+b^4)(a^4-a^3b+a^2b^2-ab^3+b^4)\right] &\ge 0 
\end{array}
Now it's not clear what I have to do next.I am stuck.
Note: My book doesn't teach any advanced technique for solving inequality as AM-GM ,Cauchy inequality etc.. 

Multiplying out, this is equivalent to $a^7b^2 + a^2b^7 \geq a^5b^4 + a^4b^5$, or $a^5 + b^5 \geq a^3 b^2 + a^2b^3$ (the case $ab=0$ is easy).
We can prove this using weighted AM-GM:
$$\frac{3}{5}a^5 + \frac{2}{5}b^5 \geq a^3 b^2$$
$$\frac{2}{5}a^5 + \frac{3}{5}b^5 \geq a^2 b^3$$
(Sorry, I didn't notice that you specified no AM-GM, but I'll leave this up in case others are interested.)
