# 1999 IMO Problems/Problem 2

## Problem

(Marcin Kuczma, Poland) Let $n \geq 2$ be a fixed integer.

• (a) Find the least constant $C$ such that for all nonnegative real numbers $x_1, \dots, x_n$, $\sum_{1\leq i • (b) Determine when equality occurs for this value of $C$. ## Solution The answer is $C=1/8$, and equality holds exactly when two of the $x_i$ are equal to each other and all the other $x_i$ are zero. We prove this by induction on the number of nonzero $x_i$. First, suppose that at most two of the $x_i$, say $x_a$ and $x_b$, are nonzero. Then the left-hand side of the desired inequality becomes $x_a x_b (x_a^2 + x_b^2)$ and the right-hand side becomes $(x_a + x_b)^4/8$. By AM-GM, \begin{align*} x_a x_b(x_a^2 + x_b^2) = 1/2 \left[ \sqrt{(2x_ax_b)(x_a^2 + x_b^2)} \right]^2 &\le \frac{1}{2} \left( \frac{x_a^2 + 2x_a x_b + x_b^2}{2} \right)^2 \\ &= (x_a + x_b)^2/8, \end{align*} with equality exactly when $x_a = x_b$, as desired. Now, suppose that our statement holds when at most $k$ of the $x_i$ are equal to zero. Suppose now that $k+1$ of the $x_i$ are equal to zero, for $k\ge 2$. Without loss of generality, let these be $x_1 \ge \dotsb \ge x_k \ge x_{k+1}$. We define $\[y_j = \begin{cases} x_j, & j \notin \{k, k+1\} \\ x_k + x_{k+1}, & j=k \\ 0, & j=k+1 \end{cases} ,$ and for convenience, we will denote $S= \sum_{i=1}^{k-1} x_i$. We wish to show that by replacing the $x_i$ with the $y_i$, we increase the left-hand side of the desired inequality without changing the right-hand side; and then to use the inductive hypothesis.

We note that \begin{align*} \sum_{1\le i< j \le n} x_i x_j(x_i^2 + x_j^2) ={}& \sum_{1 \le i If we replace $x$ with $y$, then $S(x_k^3 + x_{k+1}^3) + x_k^3 x_{k+1} + x_k x_{k+1}^3$ becomes $$S(x_k + x_{k+1})^3 = S(x_k + x_{k+1})^3 + 3S(x_k^2 x_{k+1} + x_k x_{k+1}^2),$$ but none of the other terms change. Since $3S > S \ge a_1 \ge a_k, a_{k+1}$, it follows that we have strictly increased the right-hand side of the equation, i.e., $$\sum_{1 \le i By inductive hypothesis, $\[\sum_{1\le i and by our choice of $y_i$, $\[\left( \sum_{i=1}^n y_i \right)^4/8 = \left( \sum_{i=1}^n x_i \right)^4/8 .$$ Hence the problem's inequality holds by induction, and is strict when there are more than two nonzero $x_i$, as desired. $\blacksquare$

## Solution 2 (elegant)

I claim that $C=\frac{1}{8}$. Let $P_2=\sum_{i=1}^nx_i^2$ and $S_2=\sum_{1\leq i. Note that $\[\sum_{1\leq i Equality holds in the first ineq when all but two of the $x_i$ are zero, and this reduces to the $n=2$ case which we can easily show to be equivalent to $\left(x_1-x_2\right)^4\geq0$, so $x_1=x_2$. That is, equality holds when two $x_i$ are the same and the rest are zero.

Q.E.D.