# Difference between revisions of "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 $$2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.$$

## Solution 1

WLOG let $a \leq b \leq c$. Add $2(ab+bc+ca)$ to both sides of the inequality and factor to get: $$4(a(a+b+c)+bc) \geq (a+b+c)^2$$ $$\frac{4a\sqrt[3]{abc}+bc}{2} \geq 2\sqrt[3]{a^2b^2c^2}$$

The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete.

## Solution 2

WLOG let $a \leq b \leq c$. 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 $4ab + 3 \geq (a-b)^2$.

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 $x \geq 0$.