<s>[INST] Prove: $3(a^4+b^4+c^4)+48\ge 8(a^2b+b^2c+c^2a)$ Let $a, b, c$ - real numbers. Prove that $3(a^4+b^4+c^4)+48\ge8(a^2b+b^2c+c^2a)$
 [/INST] Consider the following inequality obtained by AM-GM inequality:
$$
2a^4+b^4+16=a^4+a^4+b^4+16\geq 4\sqrt[4]{16a^8b^4}=8ba^2
$$
Writing down similar inequalities for other pairs we get:
$$
2b^4+c^4+16\geq 8cb^2\\
2c^4+a^4+16\geq 8ac^2
$$
It is enough to sum up all these inequalities and we get the desired inequality.
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