Inequality. $a^7b^2+b^7c^2+c^7a^2 \leq 3 $ 
Let $a,b,c$ be positive real numbers such that $a^6+b^6+c^6=3$. Prove that
$$a^7b^2+b^7c^2+c^7a^2 \leq 3 .$$


Using AM-GM inequality, we get
$$
3 = \frac {(a^6 + b^6 + c^6)^2} 3 = \sum_{cyc} a^6 \frac {a^6 + 2b^6} 3 \geq \sum_{cyc} a^6\sqrt[3]{a^6 b^{12}} = a^8b^4 + b^8c^4 + c^8a^4
$$
Now, by means of Cauchy–Schwarz inequality we complete the proof
$$
a^3 \cdot a^4 b^2 + b^3 \cdot b^4 c^2 + c^3 \cdot c^4 a^2 \leq \sqrt{a^6 + b^6 + c^6}\sqrt{a^8 b^4 + b^8 c^4 + c^8 a^4} \leq 3 
$$
