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I want to show that $$ 2a_2^2b_1^2+a_2^2b_3^2-4a_1a_1b_1b_2-2a_2a_3b_1b_2-2a_1a_2b_2b_3-2a_2a_3b_2b_3+a_3^2b_1^2+2a_1^2b_2^2+a_3^2b_2^2+2a_1a_3b_2^2+a_1^2b_3^2-2a_1a_3b_1b_3 \geq 0. $$ This is what I have so far: $$ \left(2a_2^2b_1^2+a_2^2b_3^2-4a_1a_1b_1b_2-2a_2a_3b_1b_2-2a_1a_2b_2b_3-2a_2a_3b_2b_3+a_3^2b_1^2+2a_1^2b_2^2 +a_3^2b_2^2+2a_1a_3b_2^2+a_1^2b_3^2-2a_1a_3b_1b_3\right) $$ $$ \geq \left(a_2^2b_1^2+a_2^2b_3^2-4a_1a_1b_1b_2-2a_2a_3b_1b_2-2a_1a_2b_2b_3-2a_2a_3b_2b_3+a_3^2b_1^2+a_1^2b_2^2 +a_3^2b_2^2+2a_1a_3b_2^2+a_1^2b_3^2-2a_1a_3b_1b_3\right) $$ \begin{align*} &= (a_1b_2-a_2b_1)^2+(a_2b_3-a_3b_2)^2+(a_1b_3-a_3b_1)^2 + 2(a_2b_1-a_1b_2)(a_2b_3-a_3b_2) - 2a_1a_2b_1b_2 \\ &\geq 0 \\ &\iff 2(a_1b_2-a_2b_1)^2+(a_2b_3-a_3b_2)^2+(a_1b_3-a_3b_1)^2 + 2(a_2b_1-a_1b_2)(a_2b_3-a_3b_2) \geq a_1^2b_2^2+a_2^2b_1^2 \\ &\iff \left( (a_1b_2-a_2b_1)^2+(a_2b_3-a_3b_2)^2\right)^2+(a_1b_2-a_2b_1)^2+(a_1b_3-a_3b_1)^2 \geq a_1^2b_2^2+a_2^2b_1^2. \end{align*} From here, I am stuck. What can I do with the right hand side of the final inequality to get to the result? Any hints?

(I apologize that the formatting looks ugly. There are a lot of characters, so it is hard to get it aligned nicely.)

  • This is a variation on the theme. – user64494 Aug 08 '13 at 17:41
  • @user64494: Hm, sorry I do not see the connection. –  Aug 08 '13 at 17:59
  • @ 5space: The LHS of the inequality is a homogenous polynomial of degree 4 in 6 variables. If all these are equal, then the equality 0=0 is obtained. Up to the cited link, consider the case of the opposite inequality and so on. Don't hesitate to ask for further explanations in need. – user64494 Aug 08 '13 at 18:12

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