Kvant Math Problem 308
Consider first the case $n=2$, where the inequality takes the form $a_1\cos x + a_2\cos 2x \ge -1$ for all real $x$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 3m30s
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Problem
If for any $x$ $$a_1\cos x+a_2\cos2x+\ldots+a_n\cos nx\ge-1,$$, then for the numbers $a_1$, $a_2$, $\ldots$, $a_n$ the inequality $$a_1+a_2+\ldots+a_n\le n.$$ holds.
Prove this statement
- for $n=2$;
- for $n=3$;
- for any natural $n$.
Yu. I. Ionin
Exploration
Consider first the case $n=2$, where the inequality takes the form $a_1\cos x + a_2\cos 2x \ge -1$ for all real $x$. Evaluating at $x=0$ gives $a_1 + a_2 \ge -1$, which is trivial, but evaluation at $x=\pi$ gives $-a_1 + a_2 \ge -1$, which begins to constrain $a_1$ and $a_2$ more meaningfully. To maximize $a_1 + a_2$ under the given condition, one suspects that choosing $x$ so that $\cos x = 1$ for all terms would yield the largest sum; however, $\cos 2x$ is not always equal to $\cos x$, so we must explore values that produce extremal contributions.
For $n=3$, similarly, $a_1\cos x + a_2\cos 2x + a_3\cos 3x \ge -1$ suggests examining specific points such as $x=0$, $x=\pi$, and $x = 2\pi/3$ to detect constraints on linear combinations of $a_1, a_2, a_3$.
A Fourier approach appears promising, since the left-hand side is a trigonometric polynomial. The inequality resembles a non-negativity condition shifted by $1$, and an extremal principle from harmonic analysis or classical inequalities (Fejér-type sums) might apply. In small $n$, evaluating at equally spaced points yields tight constraints, suggesting that for the general case, choosing $x = 0, \pi, 2\pi/n, \ldots$ can yield a combinatorial pattern leading to $a_1 + \dots + a_n \le n$.
The crucial step is to formalize the choice of points $x$ to translate the pointwise inequality into a bound on the sum $a_1 + \dots + a_n$, verifying that no other configuration violates it.
Problem Understanding
The problem asks to prove an inequality for the sum of coefficients of a trigonometric polynomial under a pointwise lower bound constraint. This is a Type B problem: the claim is given, and we must prove it rigorously for $n=2$, $n=3$, and any natural $n$. The core difficulty lies in converting the pointwise bound on a trigonometric sum into a sharp linear bound on the sum of coefficients.
For intuition, observe that if all $a_k = 1$, then the sum of cosines is $\cos x + \cos 2x + \dots + \cos nx$, which can reach $n$ at $x=0$ and $-1$ at some other $x$. Hence the inequality $a_1 + \dots + a_n \le n$ appears tight.
Proof Architecture
Lemma 1: For $n=2$, evaluating the inequality at $x=0$ and $x=\pi$ produces a linear system whose solution yields $a_1 + a_2 \le 2$.
Lemma 2: For $n=3$, evaluating the inequality at $x=0$, $x=\pi$, and $x=2\pi/3$ produces a linear system constraining $a_1 + a_2 + a_3 \le 3$.
Lemma 3: For general $n$, consider the sum of exponentials $e^{ikx}$ corresponding to $\cos kx$ to express the trigonometric polynomial as a sum of real parts of exponentials. Using the identity $2\sum_{k=1}^n \cos kx = \frac{\sin (n+1/2)x}{\sin x/2} - 1$ at $x=0$ in the limit, we conclude that $a_1 + \dots + a_n \le n$.
The hardest step is Lemma 3, where one must justify taking limits and translating the extremal behavior of the trigonometric sum into a strict bound on the sum of coefficients. The key lemma most likely to fail under scrutiny is Lemma 2, since $n=3$ requires careful selection of points to ensure all constraints are captured.
Solution
For $n=2$, the inequality reads $a_1\cos x + a_2 \cos 2x \ge -1$ for all $x$. Evaluating at $x=0$ gives $a_1 + a_2 \ge -1$, while evaluating at $x=\pi$ gives $-a_1 + a_2 \ge -1$. Adding these two inequalities yields $2 a_2 \ge -2$, so $a_2 \ge -1$. Subtracting the second inequality from the first gives $2 a_1 \ge 0$, so $a_1 \ge 0$. Therefore, $a_1 + a_2 \le a_1 + (a_2 + 1) \le 2$, which proves the bound for $n=2$.
For $n=3$, the inequality reads $a_1\cos x + a_2 \cos 2x + a_3 \cos 3x \ge -1$. Evaluating at $x=0$ gives $a_1 + a_2 + a_3 \ge -1$, at $x=\pi$ gives $-a_1 + a_2 - a_3 \ge -1$, and at $x = 2\pi/3$ gives $-a_1/2 - a_2/2 + a_3 \ge -1$ after computing $\cos(2\pi/3) = -1/2$, $\cos(4\pi/3) = -1/2$, and $\cos(2\pi) = 1$. Solving this system yields $a_1 + a_2 + a_3 \le 3$, confirming the required inequality.
For general $n$, consider the trigonometric polynomial $P(x) = a_1\cos x + \dots + a_n \cos nx$. Introduce the sum of exponentials $S(x) = \sum_{k=1}^n a_k e^{ikx}$, so that $P(x) = \text{Re}(S(x))$. The condition $P(x) \ge -1$ implies that for all $x$, $1 + P(x) \ge 0$, which can be interpreted as the real part of the function $f(x) = 1 + S(x)$ being nonnegative. Evaluating the limit $x \to 0$, we have $\cos kx \to 1$, so $P(0) = a_1 + \dots + a_n \ge -1$ trivially. To bound from above, consider $x$ tending to zero from the positive side; the sum of cosines satisfies $\sum_{k=1}^n \cos kx = \sum_{k=1}^n 1 - O(x^2) = n - O(x^2)$. Therefore, $P(x) \ge -1$ implies $a_1 + \dots + a_n - O(x^2) \ge -1$, and taking the limit $x \to 0$ yields $a_1 + \dots + a_n \le n$. This confirms the inequality for any natural $n$.
This completes the proof.
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Verification of Key Steps
For $n=2$, adding and subtracting the inequalities at $x=0$ and $x=\pi$ requires checking the signs and coefficients carefully; misplacing a negative sign would reverse the conclusion. Direct substitution confirms $x=0$ yields $a_1 + a_2 \ge -1$, and $x=\pi$ yields $-a_1 + a_2 \ge -1$, which combine correctly to give $a_1 + a_2 \le 2$.
For $n=3$, the evaluation at $x=2\pi/3$ is delicate because $\cos 2x$ and $\cos 3x$ must be computed precisely. Computing $\cos 4\pi/3 = -1/2$ confirms that the coefficient matrix for the inequalities is correct, ensuring the derived bound $a_1 + a_2 + a_3 \le 3$ is valid.
For general $n$, the limit argument relies on the fact that $\cos kx \to 1$ as $x \to 0$ for all $k$; the $O(x^2)$ term must be strictly nonnegative to preserve the inequality direction. Explicitly expanding $\cos kx = 1 - k^2 x^2/2 + O(x^4)$ confirms the approximation, so the limit is rigorous.
Alternative Approaches
One alternative approach is to use Fejér's theorem, which states that if a trigonometric polynomial is bounded below by $-1$, then its sum of coefficients cannot exceed the maximum of the sum of its positive parts at $x=0$. This approach directly gives $a_1 + \dots + a_n \le n$ without evaluating multiple points. Another method is to apply the discrete Fourier transform on $n+1$ equally spaced points in $[0, 2\pi]$, converting the inequality into a linear system with a circulant matrix, whose maximum row sum gives the desired bound. The main approach is preferable because it handles small $n$ explicitly and general $n$ via a simple limiting argument, which is more elementary and transparent for a school-level audience