Prove: $3(a^4+b^4+c^4)+48\ge 8(a^2b+b^2c+c^2a)$ [closed]
Solution 1:
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.