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Let $a$, $b$, $c$ be non-negative real numbers such that $ab + bc + ca = 3$. Suppose that \[a^3 b + b^3 c + c^3 a + 2abc(a + b + c) = \frac{9}{2}.\] What is the minimum possible value of $ab^3 + bc^3 + ca^3$?

This is an HMMT problem, with a solution here.

However, it is unnatural to start with $$ab(b+c-2a)^2+bc(c+a-2b)^2+ca(a+b-2c)^2\ge0.$$ So I want to find a solution that is easier to come up with.

Does $pqr$ method help here?

youthdoo
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  • 1/ Looking at the source, it asks for the minimum value. The AOPS solution is identical to the source, and states $\sum ab^3 \geq 18$. $\quad$ 2/ That seems to be the unique non-negative real solution (with permutations). – Calvin Lin Oct 25 '24 at 12:37
  • About $ab(b+c-2a)^2+bc(c+a-2b)^2+ca(a+b-2c)^2\ge0$: I learned this trick from @d8g3n1v9 in this question. – River Li Oct 25 '24 at 13:12
  • @RiverLi But why do we need an SOS in this form? What if it was$$\sum_{\rm cyc}\left(a^2+t_1b^2+t_2c^2+t_3ab+t_4ac+t_5bc\right)^2\ge0?$$ – youthdoo Oct 25 '24 at 14:45
  • @youthdoo Because in some approach, we need to prove that $4(a^3b + b^3c + c^3a) + ab^3 + bc^3 + ca^3 + f(a,b,c) \ge 0$. And $ab(b+c-2a)^2+bc(c+a-2b)^2+ca(a+b-2c)^2\ge0$ gives $4(a^3b + b^3c + c^3a) + ab^3 + bc^3 + ca^3 + g(a,b,c) \ge 0$. Then it suffices to prove that $f(a,b,c) \ge g(a,b,c)$. – River Li Oct 25 '24 at 14:50
  • It is HMMT 2017: https://hmmt-archive.s3.amazonaws.com/tournaments/2017/feb/guts/solutions.pdf – River Li Oct 29 '24 at 02:21
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    Yes. I said that in my question statement. @RiverLi – youthdoo Oct 31 '24 at 04:12

1 Answers1

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Remark. Here is my proof. We use the trick in my answers: P1, P2, etc.

We have \begin{align*} &ab^3 + bc^3 + ca^3 - 18\\ ={}& ab^3 + bc^3 + ca^3 - 18\\ &\qquad + 4\left(a^3b + b^3c + c^3a + 2abc(a + b + c) - \frac92\right)\\ &\qquad -24 (ab + bc + ca - 3)\\ ={}& 4(a^3b + b^3c + c^3a) + ab^3 + bc^3 + ca^3\\ &\qquad + 8abc(a + b + c) - 24(ab + bc + ca) + 36\\ ={}& ab(2a - b)^2 + bc(2b - c)^2 + ca(2c - a)^2 + 4(ab + bc + ca - 3)^2\\ \ge{}& 0. \end{align*}

River Li
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