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Here is question 1 from USAMO 2018 Q1 (held in April):

Let $a,b,c$ be positive real numbers such that $a+b+c = 4 \sqrt[3]{abc}$. Prove that: $$2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2$$

This question is on symmetric polynomials. I recall as many facts as I can think of regarding inequalities. (This test was closed book).

  • AM-GM inequality: $a + b + c \geq 3 \sqrt[3]{abc}$ so the equality there is strange.
  • quadratic mean inequality suggests $3(a^2 + b^2 + c^2) > a + b + c $.
  • The $\min(a^2,b^2,c^2)$ on the left side makes things difficult since we can't make it smaller.
  • I am still looking for other inequalities that might work.

It's tempting to race through this problem with the first solution that comes to mind. I'm especially interested in some kind of organizing principle.

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  • $\begingroup$ @MartinR the exam took place April 18. $\endgroup$ – cactus314 May 16 '18 at 13:25
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Since our inequality and condition are symmetric and homogeneous,

we can assume that $abc=1$ and $a\geq b\geq c$.

Thus, $a+b+c=4$ and we need to prove that $$2(ab+ac+bc)\geq a^2+b^2-3c^2$$ or $$(a+b+c)^2\geq2(a^2+b^2-c^2)$$ or $$8\geq a^2+b^2-c^2$$ or $$8\geq a^2+b^2-(4-a-b)^2$$ or $$8\geq a^2+b^2-(16+a^2+b^2-8a-8b+2ab)$$ or $$12-4(a+b)+ab\geq0$$ or $$12-4(4-c)+\frac{1}{c}\geq0$$ or $$(2c-1)^2\geq0.$$ Done!

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  • $\begingroup$ And the case $$a\le b\le c$$? $\endgroup$ – Dr. Sonnhard Graubner May 16 '18 at 12:45
  • $\begingroup$ In your case the solution is the same. Finally, we'll get $(2a-1)^2\geq0.$ This inequality is symmetric and we can assume that $a\geq b\geq c$. $\endgroup$ – Michael Rozenberg May 16 '18 at 12:46
  • $\begingroup$ Are these iff statements? Are you sure you can go from $(2c-1)^2 \geq 0$ to the inequality $2(ab+ac+bc) \geq a^2 + b^2 - 3c^2$ ? $\endgroup$ – cactus314 May 16 '18 at 12:47
  • $\begingroup$ Yes, of course. See my solution. I am ready to explain it more. $\endgroup$ – Michael Rozenberg May 16 '18 at 12:53
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    $\begingroup$ Very interesting equality case (under the assumption $a \geq b \geq c$): $$\left(a,b,c\right)=\left(\frac{7+\sqrt{17}}{4},\frac{7-\sqrt{17}}{4},\frac{1}{2}\right)\lambda\,,$$ where $\lambda \geq 0$. $\endgroup$ – Batominovski May 16 '18 at 12:59
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Without loss of generality, let $a\le b=ax\le c=axy, x\ge 1, y\ge 1$.

Then: $$a+b+c = 4 \sqrt[3]{abc} \Rightarrow a+ax+axy=4\sqrt[3]{a(ax)(axy)} \Rightarrow 1+x+xy=4x^{\frac23}y^{\frac13} \qquad (1)$$ Also: $$a+b+c = 4 \sqrt[3]{abc} \Rightarrow a^2+b^2+c^2=16\sqrt[3]{(abc)^2}-2(ab+bc+ca) \qquad (2)$$ Plugging $(2)$ and then $(1)$ into the given inequality: $$2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2 \overbrace{\Rightarrow}^{(2)}\\ 2(ab+bc+ca)+4a^2\ge 16\sqrt[3]{(abc)^2}-2(ab+bc+ca) \Rightarrow \\ (a(ax)+(ax)(axy)+(axy)a)+a^2\ge 4\sqrt[3]{(a(ax)(axy))^2} \Rightarrow\\ x+x^2y+xy+1\ge 4x\sqrt[3]{xy^2} \overbrace{\Rightarrow}^{(1)}\\ 4x^{\frac23}y^{\frac13}+x^2y\ge 4x^{\frac43}y^{\frac23} \Rightarrow\\ \left(x^{\frac23}y^{\frac13}\right)^2-4\left(x^{\frac23}y^{\frac13}\right)+4\ge 0 \Rightarrow \\ \left(x^{\frac23}y^{\frac13}-2\right)^2\ge 0.$$

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$ 2(ab+bc+ca)$+$ 4\min( a^2,b^2,c^2)$=$ 4(ab+bc+ca)+ 4\min(a^2,b^2,c^2)-2(ab+bc+ca)$

Now, suppose $a≤b≤c$ without losing generality. Hence,

$ 2(ab+bc+ca)$+$ 4\min( a^2,b^2,c^2)$

=$ 4(ab+bc+ca)+ 4a^2-2(ab+bc+ca)$

= $4((a+b+c)a+bc)-2(ab+bc+ca)$

=$4(4a(abc)^{1/3})+bc)-2(ab+bc+ca)≥ 16(abc)^{2/3}-2(ab+bc+ca) =(a+b+c)^2-2(ab+bc+ca)=a^2+b^2+c^2$

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