2018 USAJMO Problems/Problem 2

Problem

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

Video Solution

https://www.youtube.com/watch?v=eKIiaYNWUzM&t=5s

Solution 1

WLOG let $a \leq b \leq c$ . Add $2(ab+bc+ca)$ to both sides of the inequality and factor to get: $4(a(a+b+c)+bc) \geq (a+b+c)^2$ By substituting $a+b+c=4\sqrt[3]{abc}$ , we get: $\frac{4a\sqrt[3]{abc}+bc}{2} \geq 2\sqrt[3]{a^2b^2c^2}$ The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete.

Solution 2

WLOG let $a \geq b \geq c$ . Note that the equations are homogeneous, so WLOG let $c=1$ . Thus, the inequality now becomes $2ab + 2a + 2b + 4 \geq a^2 + b^2 + 1$ , which simplifies to $2(a+b) + 3 \geq (a-b)^2$ .

Now we will use the condition. Letting $x=a+b$ and $y=a-b$ , we have $x+1=\sqrt[3]{16(x^2-y^2)} \implies 16y^2=-x^3+13x^2-3x-1$ .

Plugging this into the inequality, we have $2x+3 \geq \frac{1}{16}(-x^3+13x^2-3x-1) \implies x^3-13x^2+35x+49 = (x-7)^2(x+1) \geq 0$ , which is true since $x \geq 0$ .

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Solution 3

https://wiki-images.artofproblemsolving.com//6/69/IMG_8946.jpg

-srisainandan6

2018 USAJMO (Problems • Resources)
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