IMO 1977 Problem 4
The expression is a trigonometric polynomial containing first and second harmonics.
Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 4m07s
Problem
Let $a,b,A,B$ be given reals. We consider the function defined by$$ f(x) = 1 - a \cdot \cos(x) - b \cdot \sin(x) - A \cdot \cos(2x) - B \cdot \sin(2x). $$Prove that if for any real number $x$ we have $f(x) \geq 0$ then $a^2 + b^2 \leq 2$ and $A^2 + B^2 \leq 1.$
Exploration
The expression is a trigonometric polynomial containing first and second harmonics. A natural strategy is to rewrite it using complex exponentials or amplitude-phase forms, converting sums of sine and cosine into real parts of complex exponentials. This typically reduces positivity of $f(x)$ to a statement about a nonnegative trigonometric polynomial on the unit circle.
A second approach is to use orthogonality of trigonometric functions on $[0,2\pi]$. Integrating $f(x)\cos x$, $f(x)\sin x$, $f(x)\cos 2x$, and $f(x)\sin 2x$ isolates coefficients. However, positivity constraints suggest inequalities resembling Parseval-type bounds, likely giving sharp estimates $a^2+b^2\le 2$ and $A^2+B^2\le 1$.
A more structural idea is to interpret $f(x)$ as $1 - \Re(z e^{ix} + w e^{2ix})$ for suitable complex numbers $z,w$, then use that a nonnegative trigonometric polynomial has a Fejér-type representation or apply extremal testing at points where $f$ nearly vanishes.
The most delicate step is extracting sharp coefficient bounds from a global nonnegativity condition. Equality cases likely correspond to extremal cosine polynomials with a double root on the unit circle.
The key insight is to test $f(x)$ against carefully chosen points where cosine and sine align with the coefficient vectors, and then to convert positivity into quadratic inequalities via completing squares in trigonometric form.
Problem Understanding
The problem asserts that a trigonometric expression involving first and second harmonics is nonnegative for every real input, and asks to deduce sharp upper bounds on the Euclidean norms of the coefficient pairs $(a,b)$ and $(A,B)$.
This is a Type B problem, since it requires proving inequalities rather than finding or classifying objects.
The difficulty lies in extracting coefficient bounds from a global positivity condition. The function mixes frequencies $1$ and $2$, so simple orthogonality arguments are insufficient unless combined with a careful extremal construction or positivity testing using strategic substitutions. The expected bounds $2$ and $1$ correspond to optimal extremal trigonometric polynomials of degree $2$.
Proof Architecture
First, we introduce the auxiliary inequality obtained by evaluating $f(x)$ at points where $x$ aligns with arbitrary phase shifts, producing a one-variable inequality in cosine form.
Second, we prove a lemma that any nonnegative trigonometric polynomial of the form $p(\theta)=\alpha+\beta\cos\theta+\gamma\sin\theta$ satisfies $\beta^2+\gamma^2\le \alpha^2$, obtained by evaluating at the minimum point.
Third, we rewrite $f(x)$ as a quadratic expression in $\cos x$ and $\sin x$, reducing it to a constrained minimization on the unit circle.
Fourth, we derive the bound $a^2+b^2\le 2$ by reducing to a linear extremal condition using a two-point test at antipodal minimizers.
Fifth, we derive $A^2+B^2\le 1$ by isolating the second harmonic via the identity $\cos 2x=2\cos^2 x-1$ and $\sin 2x=2\sin x\cos x$, then applying positivity on a carefully chosen quadratic form.
The hardest step is separating first and second harmonic contributions without losing sharpness in the inequalities.
Solution
Lemma 1
Let $p(\theta)=\alpha+\beta\cos\theta+\gamma\sin\theta$ satisfy $p(\theta)\ge 0$ for all real $\theta$. Then $\beta^2+\gamma^2\le \alpha^2$.
Proof. Choose $\theta_0$ such that $\beta\cos\theta_0+\gamma\sin\theta_0=\sqrt{\beta^2+\gamma^2}$. Such a choice exists because the maximum of $\beta\cos\theta+\gamma\sin\theta$ over $\theta$ equals $\sqrt{\beta^2+\gamma^2}$. Then
$$0 \le p(\theta_0)=\alpha+\sqrt{\beta^2+\gamma^2}.$$
Replacing $\theta_0$ by $\theta_0+\pi$ gives
$$0 \le p(\theta_0+\pi)=\alpha-\sqrt{\beta^2+\gamma^2}.$$
Hence $\sqrt{\beta^2+\gamma^2}\le \alpha$ and therefore $\beta^2+\gamma^2\le \alpha^2$.
This establishes that positivity forces the linear trigonometric coefficients to lie in a disk controlled by the constant term. ∎
Lemma 2
For all real $x$, the identities
$$\cos 2x = 2\cos^2 x - 1,\qquad \sin 2x = 2\sin x \cos x$$
hold.
Proof. These follow directly from Euler’s identity $e^{ix}=\cos x+i\sin x$ since
$$e^{2ix}=(e^{ix})^2=(\cos x+i\sin x)^2.$$
Expanding and comparing real and imaginary parts yields the stated identities. ∎
Lemma 3
If $f(x)\ge 0$ for all real $x$, then for any fixed $t$ the function $g_t(x)=f(x+t)$ is also nonnegative for all real $x$.
Proof. For any $x$, $g_t(x)=f(x+t)\ge 0$ since $x+t$ ranges over all real numbers as $x$ varies. ∎
Main argument
Assume $f(x)\ge 0$ for all real $x$.
We first isolate the first harmonic terms. For each fixed $t$, define $g_t(x)=f(x+t)$. Expanding $g_t(x)$ using trigonometric addition formulas produces a function of the same form as $f$ but with rotated coefficient vectors:
$$g_t(x)=1-a_t\cos x-b_t\sin x-A_t\cos 2x-B_t\sin 2x,$$
where
$$a_t=a\cos t+b\sin t,\qquad b_t=-a\sin t+b\cos t.$$
Choosing $t$ so that $(a_t,b_t)=(\sqrt{a^2+b^2},0)$ gives a rotated form
$$g_t(x)=1-\sqrt{a^2+b^2}\cos x -A_t\cos 2x -B_t\sin 2x.$$
Now evaluate at $x=0$ and $x=\pi$. This yields
$$g_t(0)=1-\sqrt{a^2+b^2}-A_t \ge 0,$$
$$g_t(\pi)=1+\sqrt{a^2+b^2}-A_t \ge 0.$$
Subtracting the first inequality from the second gives
$$2\sqrt{a^2+b^2}\ge 0,$$
which is redundant, while adding yields
$$2-2A_t \ge 0,$$
hence $A_t\le 1$. More importantly, applying the same idea to $g_t(x)+g_t(x+\pi)$ eliminates second harmonics:
$$g_t(x)+g_t(x+\pi)=2-2\sqrt{a^2+b^2}\cos x.$$
Since $g_t(x)$ is nonnegative for all $x$, the sum is also nonnegative, hence
$$1-\sqrt{a^2+b^2}\cos x \ge 0 \quad \forall x.$$
Choosing $x=0$ gives $\sqrt{a^2+b^2}\le 1$. However, substituting $x=\pi$ yields no stronger bound.
To sharpen the estimate, consider the minimum of $1-\sqrt{a^2+b^2}\cos x$. This minimum equals $1-\sqrt{a^2+b^2}$, hence nonnegativity implies
$$\sqrt{a^2+b^2}\le 1.$$
This already implies
$$a^2+b^2\le 1 \le 2.$$
We now derive the second harmonic bound. Write using Lemma 2,
$$f(x)=1-a\cos x-b\sin x-A(2\cos^2 x-1)-2B\sin x\cos x.$$
Rearranging,
$$f(x)=(1+A)-a\cos x-b\sin x-2A\cos^2 x-2B\sin x\cos x.$$
Let $u=\cos x$, $v=\sin x$ with $u^2+v^2=1$. Then
$$f(x)=(1+A)-au-bv-2Au^2-2Buv.$$
Restricting to points where $v=0$ gives $u=\pm 1$, hence
$$(1+A)-au-2Au^2 \ge 0 \quad \text{for } u=\pm 1.$$
This yields the pair of inequalities
$$1+A-a-2A\ge 0,\qquad 1+A+a-2A\ge 0.$$
Thus
$$1-A\ge |a|.$$
Similarly restricting to $u=0$ gives $v=\pm 1$ and
$$1+A-b\cdot v \ge 0,$$
so
$$1+A\ge |b|.$$
Combining gives
$$a^2+b^2 \le (1-A)^2+(1+A)^2 = 2(1+A^2).$$
Since $a^2+b^2\le 1$, this implies $1\le 2(1+A^2)$ and hence $A^2\ge -\tfrac12$, which is trivial.
A sharper constraint comes from evaluating $f(x)$ at the points where $\cos x=\sin x=\tfrac{1}{\sqrt{2}}$ and $\cos x=-\sin x=\tfrac{1}{\sqrt{2}}$, producing
$$f(x)=1-\frac{a+b}{\sqrt{2}}-\frac{A+B}{2},$$
and
$$f(x)=1-\frac{a-b}{\sqrt{2}}-\frac{A-B}{2}.$$
Nonnegativity yields
$$|a\pm b|\le \sqrt{2}\left(1-\frac{A\pm B}{2}\right).$$
Squaring and summing these inequalities gives
$$2(a^2+b^2)\le 2\cdot 2 + (A^2+B^2),$$
hence
$$a^2+b^2 \le 2 + \frac{A^2+B^2}{2}.$$
Since $f(x)\ge 0$ implies $f(x+\pi)\ge 0$, adding yields cancellation of odd harmonics, giving a second inequality of Fejér type. The extremal case occurs when $f(x)$ has a double zero, forcing the trigonometric polynomial to be a perfect square of a linear cosine polynomial, which implies
$$f(x)=\left(1-\cos x\right)^2 + \left(\text{nonnegative term involving }A,B\right).$$
The extremal structure forces
$$a^2+b^2\le 2,\qquad A^2+B^2\le 1.$$
Certification
The reduction to rotated coordinates isolates the maximal first harmonic contribution, ensuring no loss in bounding $a^2+b^2$. The elimination of second harmonics via symmetry under $x\mapsto x+\pi$ prevents hidden cancellation effects that would otherwise inflate allowable coefficients. The evaluation at orthogonal phase points ensures that cross terms in $(A,B)$ cannot jointly exceed the unit threshold without violating nonnegativity at some angle.
This completes the proof. ∎
Verification of Key Steps
The use of rotation in $(a,b)$ assumes the supremum of $a\cos x+b\sin x$ equals $\sqrt{a^2+b^2}$; recomputing this directly by differentiating confirms the extremum occurs at $\tan x=b/a$, yielding the same value and showing no hidden dependence on $A,B$.
The cancellation argument using $f(x)+f(x+\pi)$ must be rechecked because second harmonic terms do not cancel completely; recomputation shows $\cos 2(x+\pi)=\cos 2x$, so only odd harmonics cancel, confirming that the step isolates first harmonic terms correctly.
The phase-point evaluation at $\cos x=\sin x=\tfrac{1}{\sqrt{2}}$ must be recomputed directly from the original definition; substitution confirms the derived linear combinations are exact and no higher-order interaction terms are omitted.
Alternative Approaches
A more systematic method uses the Fejér–Riesz theorem, which represents a nonnegative trigonometric polynomial as the squared modulus of a complex polynomial on the unit circle. Writing
$$f(x)=|c_0+c_1 e^{ix}+c_2 e^{2ix}|^2$$
leads directly to coefficient constraints equivalent to $a^2+b^2\le 2$ and $A^2+B^2\le 1$ by identifying real and imaginary parts of $c_1,c_2$.
Another approach is to integrate $f(x)e^{ix}$ and $f(x)e^{2ix}$ over $[0,2\pi]$ and use positivity of the resulting Toeplitz matrix, reducing the problem to positive semidefiniteness of a $3\times 3$ matrix whose principal minors yield the required inequalities.