<s>[INST] Prove that $\sum\limits_{cyc}(4a^6+5a^5b)\geq\frac{(a+b+c)^6}{27}$ Let $a$, $b$ and $c$ be real numbers. Prove that:
$$4(a^6+b^6+c^6)+5(a^5b+b^5c+c^5a)\geq\frac{(a+b+c)^6}{27}$$
I tried SOS, uvw and more, but without success.
 [/INST] The following proof is true if they are non-negative reals. 
$4(x^3+y^3+z^3)+15xyz\geq (x+y+z)^3$- this directly follows from Schur. So $$4(a^6+b^6+b^6)+5(a^5b+b^5c+c^5a)\geq 4(a^6+b^6+c^6)+15a^2b^2c^2\geq (a^2+b^2+c^2)^3\geq\dfrac{(a+b+c)^6}{27}$$
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