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Let $a,b\in[-1,1]$,

then prove or disprove:

$$f(a,b)=3a^3b+3ab^3+18a^2b+18ab^2+12a^3+12b^3+40a^2+40b^2+64ab\ge 0$$

My try: Since \begin{align*} f(a,b)&=3a^3b+3ab^3+18a^2b+18ab^2+12a^3+12b^3+40a^2+40b^2+64ab\\ &=3ab(a^2+b^2)+18ab(a+b)+12(a^3+b^3)+40(a^2+b^2)+64ab \end{align*} if $ab\ge 0$

then \begin{align*}&f(a,b)\ge 9ab(a^2+b^2)+18ab(a+b)+12(a^3+b^3)+40(a^2+b^2)+52ab\\ &=9ab(a^2+b^2)+18ab(a+b)+12(a+b)(a^2+b^2-ab)+40(a^2+b^2)+52ab\\ &=9ab(a^2+b^2)+6(a+b)(2a^2+2b^2+ab)+40(a^2+b^2)+52ab\\ &\ge0 \end{align*}

but for the other case, $ab\le 0$, I can't proceed.

This problem is a follow up from this.

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1 Answers1

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for $ab<0$, WLOG,let $a>0,-c=b<0 \to 1 \ge c>0$ so the inequality becomes:

$-3a^3c-3ac^3-18a^2c+18ac^2+12a^3-12c^3+40a^2+40c^2-64ac\ge 0$

$12c^2 \ge 12c^3,3ac^3 \ge 3ac^3,9ac^2+9a^3 \ge 18a^2c,3a^3 \ge 3a^3c,40a^2+28c^2\ge 64ac $

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