0

For $a,b,c \geq 0$ show that \begin{eqnarray*} 3(a^4+b^4+c^4) +abc(a+b+c) \geq 4(a^2b^2+b^2c^2+c^2a^2). \end{eqnarray*} I have managed to find a proof of this, but it is gruesome & I will post my solution shortly. I am really interested in seeing some more elegant solutions.

I came across this question while trying to solve this question prove this inequality by $abc=1$

Any comments & slick solutions are welcome.

Michael Rozenberg
  • 1
Donald Splutterwit
  • 37,542
  • I think that you have to prove this inequality by setting $a+b+c=1$. – xpaul Apr 27 '18 at 01:27
  • The issue with this problem is that $abc(a+b+c) \leq a^2b^2 + c^2b^2 + c^2a^2$ for all $a,b,c \geq 0$, so you cannot deal with terms separately on the LHS. Rather, you need some inequality that gives both terms on the left simultaneously. – Sarvesh Ravichandran Iyer Apr 27 '18 at 01:28
  • @астонвіллаолофмэллбэрг Your right, one is inclined to split it into two inequalities, but the sign is reversed in the inequality you state. How do you incorporate this into the other inequality ? ... in a cool way. – Donald Splutterwit Apr 27 '18 at 01:32

2 Answers2

2

BY Schur and AM-GM we obtain: $$\sum_{cyc}(3a^4+a^2bc-4a^2b^2)=$$ $$=\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+\sum_{cyc}(a^4+b^4+a^3b+ab^3-4a^2b^2)\geq0.$$

Michael Rozenberg
  • 1
0

It is clear that \begin{eqnarray*} 3(a^2(2a-b-c)^2+b^2(2b-a-c)^2+c^2(2c-a-b)^2)+ 12(ab(a-b)^2+bc(b-c)^2+ca(a-c)^2)+ a^2(b-c)^2+b^2(c-a)^2+c^2(a-b)^2 \geq 0. \end{eqnarray*} Now rearrange and divide by $4$ and the result follows.

Alternatively cut & paste the following into reduce (3*(a^2*(2*a-b-c)^2+b^2*(2*b-a-c)^2+c^2*(2*c-a-b)^2)+ 12*(ab(a-b)^2+bc(b-c)^2+ca(a-c)^2)+ a^2*(b-c)^2+b^2*(c-a)^2+c^2*(a-b)^2);

Donald Splutterwit
  • 37,542