Given two natural numbers $m, n$ so that $m> n.$ Prove that there exist any real numbers $x$ so that $2\sin nx\cos mx\geq 1$

Question

Given two natural numbers $m, n$ so that $m> n.$ Prove that there exist any real numbers $x$ so that (unsolved in a case..) $$2\sin nx\cos mx\geq 1$$ Source: StackMath/@RiverLi_ Prove: $(\forall m, n\in\Bbb N_{>0})(\exists x\in\Bbb R)$ s. t. $2\sin n x \cos m x \ge 1$ .

I couldn't find a concrete value of $x,$ but I succeeded at proving the problem by using Intermediate value theorem (My friend has taught me a solving plan..). Let $x$ on the interval $I= \left [ -\frac{5}{6}\pi, \frac{\pi}{6} \right ],$ we call $f\left ( x \right ):=\frac{1}{2\sin x}, g\left ( x \right ):=\cos\alpha x, \alpha:=\frac{m}{n}> 1,$ I devide my problem into 2 cases as followings

When $1< \alpha< \frac{6}{5},$ we have $f, g$ continuous both and $f\left ( -\frac{3}{4}\pi \right )\geq g\left ( -\frac{3}{4}\pi \right ),$ by Intermediate value theorem $\exists x\in I$ so that $f\left ( x \right )= g\left ( x \right )$

When $\alpha\geq\frac{6}{5},$ we have $f, g$ continuous and $\exists x\in I$ so $g\left ( x \right )= -1\Rightarrow\exists x\in I$ so that $f\left ( x \right )= g\left ( x \right )$

Conclusion. $$\forall\alpha= \frac{m}{n}> 1, \exists x\in I: f\left ( x \right )= g\left ( x \right )\Leftrightarrow\forall m, n, \exists x\in I: 2\sin x\cos\alpha x= 1$$ $$\Rightarrow\forall m, n, \exists x\in \mathbb{R}: 2\sin nx\cos mx\geq 1$$ I'm waiting for @RiverLi's complete solution and want to see such a value of $x$ for my proof, thanks

Answer 1

Extended comment ($a=\alpha$).

Your proof is hard to follow. First, you have not specified a point $x\in I$, s.t. $f(x)<g(x)$. Second one cannot apply IVT on the interval $I=\left[-\frac{5\pi}6, \frac{\pi}{6}\right]$ because the function $f(x)$ is not continuous on this interval. Third the claim that there exists a point with $g(x)=-1$ in the chosen interval is not proven.

Below I will try to remedy these shortcomings. First we define the interval to be $I=\left[-\frac{3\pi}4, -\frac{\pi}{3a}\right]$. Clearly both functions are continuous on this interval and $f\left(-\frac{\pi}{3a}\right)<g\left(-\frac{\pi}{3a}\right)=\frac12.$ So we need only to show that there is a point $x\in I$, s.t. $f(x)>g(x)$.

Similarly as in you proof we let the point be $x=-\frac{3\pi}4$ for $1<a<\frac43$. For larger $a$ we can choose any point $x$ such that $\cos(a x)=-1$, $-\frac{3\pi}4\le x<-\frac{\pi}4$ (this choice is somewhat artificial and done only due to some symmetry). To find $x$ we need to resolve the following inequality for $k\in\mathbb Z$: $$ -\frac{3\pi}4\le -\frac\pi a(2k+1)<-\frac{\pi}4\iff \frac a4<2k+1\le\frac{3a}4.\tag1 $$ The essence here is: for $a>\frac43$ the inequality $(1)$ always has at least one solution. Observe that $k=0$ is the only solution of $(1)$ for $a<4$ and cancels to be a solution for $a>4$. This however does not harm as we can take $k=1$ for $4<a<\frac{20}3$. For higher values of $a$ multiple solutions for $k$ are possible (though smaller $k$ repeatedly cancel to be the solutions).

Answer 2

The following approach comes from looking at the graphs of the involved functions.

Given $m>n>0$ put $$m+n=:p\geq3,\qquad m-n=:q\geq1\ .\tag{1}$$ We have to study the function $$f(t):=2 \sin(nt)\cos(mt)=\sin(pt)-\sin (qt)$$ on the circle ${\mathbb T}:={\mathbb R}/(2\pi{\mathbb Z})$. After scaling the $t$-axis appropriately we may assume ${\rm gcd}(p,q)=1$. The function $t\mapsto\sin(pt)$ has $p$ equidistant local maxima (value $=1$) on ${\mathbb T}$, color them red, and the function $t\mapsto-\sin(qt)$ has $q$ equidistant local maxima (value $=1$) on ${\mathbb T}$, color them blue. Denote by $\delta\geq0$ the minimal distance between a red and a blue maximal point. I claim that $$\delta\leq{\pi\over p q}\ .\tag{2}$$ Proof. When a red and a blue point coincide we have $\delta=0$. In this case it follows from Bézout's identity that there are two more red/blue pairs, with opposite order along ${\mathbb T}$, having distance ${2\pi\over pq}$. When the blue pattern of points is then rigidly rotated vs.\ the red pattern the two coinciding points are disassociating, while one pair of distance ${2\pi\over pq}$ points is going to combine. The worst case is in the middle, when the rotation angle is ${\pi\over pq}$. At that moment the $\delta$ of the situation is just covered by $(2).\quad{\square}$

We therefore have a red maximum at $\xi\in{\mathbb T}$ and a blue maximum at $\eta\in{\mathbb T}$ with $$\delta:=|\xi-\eta|\leq{\pi\over pq}\ .$$ Let $\tau$ be the point between $\xi$ and $\eta$ having distance ${q^2\over p^2+q^2}\delta$ from $\xi$ and ${p^2\over p^2+q^2}\delta$ from $\eta$. We are now going to estimate $f(\tau)$.

In the neighborhood of $t=\xi$ the graph of $t\mapsto\sin(pt)$ is congruent to the graph of $$t\mapsto\cos(pt)\geq1-{p^2\over2}t^2$$ in the neighborhood of $t=0$. It follows that $$\sin(p\tau)\geq 1-{p^2\over2}(\tau-\xi)^2\geq1-{p^2\over2}\left({q^2\over p^2+q^2}\delta\right)^2\ .$$ Similarly we have $$-\sin(q\tau)\geq 1-{q^2\over2}(\tau-\eta)^2\geq1-{q^2\over2}\left({p^2\over p^2+q^2}\delta\right)^2\ .$$ Using $(1)$ and $(2)$ we now obtain $$f(\tau)\geq2-{p^2q^2\delta^2\over 2(p^2+q^2)}\geq2-{\pi^2\over 2(p^2+q^2)}>1\ .$$

Example: When $m=3$ and $n=2$ then $p=5$, $q=1$. Our estimate then gives $f(\tau)\geq1.81$. This value is indeed realized by $f$: