# 2018 USAMO Problems/Problem 1

## Problem 1

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

## Solution

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

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

-srisainandan6

## Solution 3

Similarly to Solution 2, we will prove homogeneity but we will use that to solve the problem differently. Let $f(a,b,c)=a+b+c-4\sqrt{abc}$. Note that $f(a,b,c)=f(ka,kb,kc)$, thus proving homogeneity.

WLOG, we can scale down all variables such that the lowest one is $1$. WLOG, let this be $a=1$. We now have $1+b+c=4\sqrt{bc}$, and we want to prove $2bc+2b+2c+4\ge 1+b^2+c^2.$ Adding $2bc$ to both sides and subtracting $2b+2c$ gives us $4bc+4\ge 1+ (b+c)(b+c-2)$, or $4bc+3\ge (b+c)(b+c-2)$. Let $\sqrt{bc}=x$. Now, we have $$4x^3+3 \ge (4x-1)(4x-3)$$ $$4x^3 - 16x^2 + 16x \ge 0$$ $$4x^2 - 16 + 16 \ge 0$$ $$4(x-2)^2 \ge 0$$ By the trivial inequality, this is always true. Since all these steps are reversible, the proof is complete. ~SigmaPiE