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

Let $a, b, c$ - real numbers. Prove that $3(a^4+b^4+c^4)+48\ge8(a^2b+b^2c+c^2a)$

• I think is $3(a^4+b^4+c^4)+48\ge 8(a^3b+b^3c+c^3a)$?
– user94270
Commented Oct 10, 2013 at 10:16
• I dont thinks so... Commented Oct 10, 2013 at 10:18
• Hello, welcome to Math.SE. Please read meta.math.stackexchange.com/questions/9959/… and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you edit your question to include some motivation, and an explanation of your own attempts. Commented Oct 10, 2013 at 10:26
• All that is an old-fashioned math. The Mathematica command $$Minimize[3*(a^4 + b^4 + c^4) + 48 - 8*(a^2*b + a*c^2 + b^2*c), \{a, b, c\}]$$ outputs $$\{0, \{a -> 2, b -> 2, c -> 2\}\} .$$ Commented Oct 10, 2013 at 19:42
• @user64494 Mathematica isn't mathematics! Commented Oct 16, 2013 at 13:50

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.