Kvant Math Problem 1144
Let
Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m12s
Source on kvant.digital
Problem
Let $a_1$, $a_2$, $\ldots$, $a_n$ be nonnegative numbers. Which number is greater: $$\sqrt[\scriptstyle1988]{a_1^{1988}+a_2^{1988}+\ldots+a_n^{1988}}\quad\text{или}\quad \sqrt[\scriptstyle1989]{a_1^{1989}+a_2^{1989}+\ldots+a_n^{1989}}~?$$
A. I. Shekhorsky
Exploration
Let
$$X=\left(a_1^{1988}+a_2^{1988}+\cdots+a_n^{1988}\right)^{1/1988}, \qquad Y=\left(a_1^{1989}+a_2^{1989}+\cdots+a_n^{1989}\right)^{1/1989}.$$
The question asks whether $X$ or $Y$ is larger.
First test simple cases.
If $n=1$, then
$$X=Y=a_1.$$
If $a_1=a_2=1$, then
$$X=2^{1/1988},\qquad Y=2^{1/1989},$$
and since $1/1988>1/1989$,
$$X>Y.$$
If $a_1=2$, $a_2=0$, then
$$X=Y=2.$$
These examples suggest that $X\ge Y$, with equality when at most one $a_i$ is nonzero.
The expression is the $\ell_p$ norm of the vector $(a_1,\dots,a_n)$ for $p=1988$ and $p=1989$. A standard fact is that for nonnegative vectors, $\ell_p$ norms decrease as $p$ increases. Since this is exactly the statement to be proved, it must be justified directly.
A convenient route is to set
$$S=\sum_{i=1}^n a_i^{1988}.$$
If $S=0$, then all $a_i=0$ and $X=Y=0$.
Assume $S>0$ and define
$$b_i=\frac{a_i^{1988}}{S}.$$
Then $b_i\ge0$ and $\sum b_i=1$. We obtain
$$\sum a_i^{1989} = S\sum b_i a_i.$$
Since the $b_i$ form weights summing to $1$,
$$\sum b_i a_i\le \max_i a_i.$$
Also,
$$a_i^{1988}\le S,$$
hence
$$a_i\le S^{1/1988}.$$
Therefore
$$\sum a_i^{1989}\le S^{1+1/1988}=S^{1989/1988}.$$
Taking $1989$th roots gives
$$Y\le S^{1/1988}=X.$$
The crucial point is proving
$$\sum a_i^{1989}\le \left(\sum a_i^{1988}\right)^{1989/1988}.$$
The weighted-average argument above appears to establish exactly this.
Problem Understanding
We are asked to compare the two numbers
$$\left(\sum_{i=1}^n a_i^{1988}\right)^{1/1988} \quad\text{and}\quad \left(\sum_{i=1}^n a_i^{1989}\right)^{1/1989},$$
where all $a_i$ are nonnegative.
This is a Type C problem, because we must determine which quantity is larger.
The core difficulty is proving the inequality
$$\left(\sum a_i^{1988}\right)^{1/1988} \ge \left(\sum a_i^{1989}\right)^{1/1989}$$
for arbitrary nonnegative numbers.
The answer is that the first number is always at least as large as the second. Intuitively, raising the exponent places more weight on larger coordinates, and the corresponding root does not fully compensate for this effect.
Proof Architecture
Let
$$S=\sum_{i=1}^n a_i^{1988}.$$
If $S=0$, both quantities are $0$.
Assume $S>0$ and define
$$b_i=\frac{a_i^{1988}}{S}.$$
Then $b_i\ge0$ and $\sum b_i=1$.
Lemma 1. For every $i$,
$$a_i\le S^{1/1988}.$$
This follows from $a_i^{1988}\le S$.
Lemma 2. One has
$$\sum_{i=1}^n a_i^{1989} = S\sum_{i=1}^n b_i a_i.$$
This is obtained by substituting $a_i^{1988}=Sb_i$.
Lemma 3. Since the $b_i$ are nonnegative and sum to $1$,
$$\sum b_i a_i\le \max_i a_i.$$
A weighted average cannot exceed the largest entry.
Combining Lemmas 1, 2, and 3 yields
$$\sum a_i^{1989} \le S^{1989/1988}.$$
Taking $1989$th roots gives the desired comparison.
The step most likely to fail under scrutiny is the passage from Lemma 2 to the bound on $\sum a_i^{1989}$, because it uses both the weighted-average property and the estimate $\max a_i\le S^{1/1988}$.
Solution
Denote
$$X=\left(a_1^{1988}+a_2^{1988}+\cdots+a_n^{1988}\right)^{1/1988},$$
and
$$Y=\left(a_1^{1989}+a_2^{1989}+\cdots+a_n^{1989}\right)^{1/1989}.$$
We shall prove that $X\ge Y$.
Let
$$S=\sum_{i=1}^n a_i^{1988}.$$
If $S=0$, then every $a_i=0$, hence $X=Y=0$.
Assume now that $S>0$. Define
$$b_i=\frac{a_i^{1988}}{S}.$$
Then
$$b_i\ge0, \qquad \sum_{i=1}^n b_i=1.$$
Since $a_i^{1988}\le S$ for every $i$,
$$a_i\le S^{1/1988}.$$
Therefore
$$\max_i a_i\le S^{1/1988}.$$
Next,
$$\sum_{i=1}^n a_i^{1989} = \sum_{i=1}^n a_i\cdot a_i^{1988} = S\sum_{i=1}^n b_i a_i.$$
Because the numbers $b_i$ are nonnegative and their sum is $1$, the quantity $\sum b_i a_i$ is a weighted average of the numbers $a_i$. Hence
$$\sum_{i=1}^n b_i a_i\le \max_i a_i.$$
Combining this with the previous estimate gives
$$\sum_{i=1}^n a_i^{1989} \le S\cdot S^{1/1988} = S^{1+1/1988} = S^{1989/1988}.$$
Taking the $1989$th root of both sides,
$$Y = \left(\sum_{i=1}^n a_i^{1989}\right)^{1/1989} \le \left(S^{1989/1988}\right)^{1/1989} = S^{1/1988} = X.$$
Thus
$$\left(a_1^{1988}+a_2^{1988}+\cdots+a_n^{1988}\right)^{1/1988} \ge \left(a_1^{1989}+a_2^{1989}+\cdots+a_n^{1989}\right)^{1/1989}.$$
Equality holds when at most one of the numbers $a_i$ is nonzero.
Hence the greater number is
$$\boxed{\left(a_1^{1988}+a_2^{1988}+\cdots+a_n^{1988}\right)^{1/1988}}.$$
Verification of Key Steps
The first delicate step is the estimate
$$a_i\le S^{1/1988}.$$
Since
$$S=\sum_{k=1}^n a_k^{1988},$$
every summand satisfies
$$a_i^{1988}\le S.$$
All quantities are nonnegative, so taking $1988$th roots yields
$$a_i\le S^{1/1988}.$$
The second delicate step is
$$\sum b_i a_i\le \max_i a_i.$$
Let $M=\max_i a_i$. Since $a_i\le M$ for every $i$,
$$\sum b_i a_i \le \sum b_i M = M\sum b_i = M.$$
The third delicate step is the transition
$$\sum a_i^{1989} = S\sum b_i a_i.$$
Using the definition $b_i=a_i^{1988}/S$,
$$S\sum b_i a_i = S\sum \frac{a_i^{1988}}{S}a_i = \sum a_i^{1989},$$
which is an exact identity, not merely an inequality.
Alternative Approaches
A standard argument uses Hölder's inequality. Let
$$p=\frac{1989}{1988},\qquad q=1989.$$
Applying Hölder to the sequences $a_i^{1988}$ and $1$ gives
$$\sum a_i^{1988} \le \left(\sum a_i^{1989}\right)^{1988/1989} n^{1/1989}.$$
After rearrangement one obtains the monotonicity of $\ell_p$ norms. This method proves a more general theorem, but it introduces machinery stronger than the problem requires.
Another approach uses Jensen's inequality. Setting
$$b_i=\frac{a_i^{1988}}{\sum a_k^{1988}},$$
the convexity of the function $x^{1989/1988}$ yields
$$\sum b_i^{1989/1988}\le 1.$$
After rewriting this inequality in terms of the $a_i$, one arrives at
$$\sum a_i^{1989} \le \left(\sum a_i^{1988}\right)^{1989/1988}.$$
The weighted-average argument used in the main proof is shorter and relies only on elementary properties of nonnegative numbers.