# 2008 iTest Problems/Problem 70

## Problem

After swimming around the ocean with some snorkling gear, Joshua walks back to the beach where Alexis works on a mural in the sand beside where they drew out symbol lists. Joshua walks directly over the mural without paying any attention.

"You're a square, Josh."

"No, $\textit{you're}$ a square," retorts Joshua. "In fact, you're a $\textit{cube}$, which is 50% freakier than a square by dimension. And before you tell me I'm a hypercube, I'll remind you that mom and dad confirmed that they could not have given birth to a four dimension being."

"Okay, you're a cubist caricature of male immaturity," asserts Alexis.

Knowing nothing about cubism, Joshua decides to ignore Alexis and walk to where he stashed his belongings by a beach umbrella. He starts thinking about cubes and computes some sums of cubes, and some cubes of sums: \begin{align*}1^3+1^3+1^3&=3,\\1^3+1^3+2^3&=10,\\1^3+2^3+2^3&=17,\\2^3+2^3+2^3&=24,\\1^3+1^3+3^3&=29,\\1^3+2^3+3^3&=36,\\(1+1+1)^3&=27,\\(1+1+2)^3&=64,\\(1+2+2)^3&=125,\\(2+2+2)^3&=216,\\(1+1+3)^3&=125,\\(1+2+3)^3&=216.\end{align*}

Josh recognizes that the cubes of the sums are always larger than the sum of cubes of positive integers. For instance, \begin{align*}(1+2+4)^3&=1^3+2^3+4^3+3(1^2\cdot 2+1^2\cdot 4+2^2\cdot 1+2^2\cdot 4+4^2\cdot 1+4^2\cdot 2)+6(1\cdot 2\cdot 4)\\&>1^3+2^3+4^3.\end{align*}

Josh begins to wonder if there is a smallest value of n such that $(a+b+c)^3\leq n(a^3+b^3+c^3)$ for all natural numbers $a, b$, and $c$. Joshua thinks he has an answer, but doesn't know how to prove it, so he takes it to Michael who confirms Joshua's answer with a proof. What is the correct value of $n$ that Joshua found?

## Solution

Note that $(a+b+c)^3 = a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3ab^2 + 3b^2c + 3ac^2 + 3bc^2 + 6abc$. By the AM-GM Inequality, $$2a^3 + 2b^3 + 2c^3 \ge 6abc$$ $$a^3 + a^3 + b^3 \ge 3a^2b$$ $$a^3 + a^3 + c^3 \ge 3a^2c$$ $$a^3 + b^3 + b^3 \ge 3ab^2$$ $$b^3 + b^3 + c^3 \ge 3b^2c$$ $$a^3 + c^3 + c^3 \ge 3ac^2$$ $$b^3 + c^3 + c^3 \ge 3bc^2$$

Adding the inequalities results in $$6a^3 + 6b^3 + 6c^3 \ge 3a^2b + 3a^2c + 3ab^2 + 3b^2c + 3ac^2 + 3bc^2 + 6abc.$$

Since $3a^3 + 3b^3 + 3c^3 = 3a^3 + 3b^3 + 3c^3$, we can add the equation as well as all 7 inequalities to get \begin{align*} \\ 9a^3 + 9b^3 + 9c^3 &\ge a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3ab^2 + 3b^2c + 3ac^2 + 3bc^2 + 6abc \\ &\ge (a+b+c)^3 \end{align*}

The equality statement can be satisfied if we let $a = b = c$, so $n = \boxed{9}$.