Difference between revisions of "1999 IMO Problems/Problem 2"

Revision as of 19:26, 23 October 2015

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 $x_1, \dots, x_n$ ,

$\sum_{1\leq i<j \leq n} x_ix_j (x_i^2 + x_j^2) \leq C \left( \sum_{i=1}^n x_i \right)^4.$

(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 $\begin{align*} \sum_{1\le i< j \le n} x_i x_j(x_i^2 + x_j^2) ={}& \sum_{1 \le i<j \le k-1} x_i x_j(x_i^2 + x_j^2) + \sum_{1 \le i \le k-1} \left[ x_ix_k(x_i^2 + x_k^2) + x_ix_{k+1}(x_i^2 + x_{k+1}^2) \right] \\ & + x_k x_{k+1}(x_k^2 + x_{k+1}^2) \\ ={}& \sum_{1 \le i<j \le k-1} x_i x_j(x_i^2 + x_j^2) + \sum_{i=1}^{k-1} x_i^3 (x_k + x_{k+1}) + S(x_k^3 + x_{k+1}^3) \\ &+ x_k^3 x_{k+1} + x_k x_{k+1}^3 \end{align*}$ 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., $\sum_{1 \le i<j \le n} x_i x_j (x_i^2 + x_j^2) < \sum_{1 \le i<j \le n} y_i y_j (y_i^2 + y_j^2) .$ By inductive hypothesis, $\sum_{1\le i<j \le n} y_i y_j (y_i^2 +y_j^2) \le \left( \sum_{i=1}^n y_i \right)^4 / 8 ,$ 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 $S_2=\sum_{1\leq i<j\leq n}x_ix_j$ . Note that $\sum_{1\leq i<j\leq n}x_ix_j\left(x_i^2+x_j^2\right)\leq\sum_{1\leq i<j\leq n}x_ix_jP_2=P_2S_2=\frac{\left(2\sqrt{P_2\cdot2S_2}\right)^2}{8}\leq\frac{\left(P_2+2S_2\right)^2}{8}=\frac{1}{8}\left(\sum_{i=1}^nx_i\right)^4.$ 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.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

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 == Solution 2 (elegant) ==
-I claim that <math>C=\frac{1}{8}</math>. Let <math>P_2=\sum_{i=1}^nx_i^2</math> and <math>S_2=\sum_{1\leq i<j\leq n}x_ix_j</math>. Note that <cmath>\sum_{1\leq i<j\leq n}x_ix_j\left(x_i^2+x_j^2\right)\leq\sum_{1\leq i<j\leq n}x_ix_jP_2=P_2S_2=\frac{\left(2\sqrt{P_2\cdot2S_2}\right)^2}{8}\leq\frac{\left(P_2+2S_2\right)^2}{8}=\frac{1}{8}\left(\sum_{i=1}^nx_i\right)^4.</cmath> Equality holds in the first ineq when all but two of the <math>x_i</math> are zero, and this reduces to the <math>n=2</math> case which we can easily show to be equivalent to <math>\left(x_1-x_2\right)^2</math>, so <math>x_1=x_2</math>. That is, equality holds when two <math>x_i</math> are the same and the rest are zero.
+I claim that <math>C=\frac{1}{8}</math>. Let <math>P_2=\sum_{i=1}^nx_i^2</math> and <math>S_2=\sum_{1\leq i<j\leq n}x_ix_j</math>. Note that <cmath>\sum_{1\leq i<j\leq n}x_ix_j\left(x_i^2+x_j^2\right)\leq\sum_{1\leq i<j\leq n}x_ix_jP_2=P_2S_2=\frac{\left(2\sqrt{P_2\cdot2S_2}\right)^2}{8}\leq\frac{\left(P_2+2S_2\right)^2}{8}=\frac{1}{8}\left(\sum_{i=1}^nx_i\right)^4.</cmath> Equality holds in the first ineq when all but two of the <math>x_i</math> are zero, and this reduces to the <math>n=2</math> case which we can easily show to be equivalent to <math>\left(x_1-x_2\right)^4\geq0</math>, so <math>x_1=x_2</math>. That is, equality holds when two <math>x_i</math> are the same and the rest are zero.
 Q.E.D.

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