Showing $\frac{[\sin y \ (x+\frac{1}{x}-2\cos y)]^2}{x^2+6} \leq 1$ for $x>1$

Question

I was trying to solve Ex 21 Chapter 5 from Stevenhagen, Number Theory and I came up into this trigonometric inequality which would be:

$$\frac{[\sin y \ (x+\frac{1}{x}-2\cos y)]^2}{x^2+6} \leq 1 \quad \text{ for } x>1$$ I tried to maximize the multivariate function but couldn't conclude. Does anyone have any ideas/hints on how to solve this?

Answer 1

Let $$p := x + \frac{15}{4x}.$$

By AM-GM, we have \begin{align*} &\left[\sin y \cdot \left(x+\frac{1}{x}-2\cos y\right)\right]^2\\[6pt] ={}& \frac{1}{p^2 - 4}\cdot (p + 2)(1 + \cos y) \cdot (p-2) (1 - \cos y) \cdot \left(x + \frac{1}{x} - 2\cos y\right)^2\\[6pt] \le{}& \frac{1}{p^2 - 4}\left[\frac{(p + 2)(1 + \cos y) + (p - 2) (1 - \cos y) + \left(x + \frac{1}{x} - 2\cos y\right) \cdot 2}{4}\right]^4\\[6pt] ={}& \frac{1}{p^2 - 4}\left(\frac{p}{2} + \frac{x}{2} + \frac{1}{2x}\right)^4\\[6pt] ={}& \frac{1}{p^2 - 4}\left(x + \frac{19}{8x}\right)^4. \end{align*}

It suffices to prove that $$(x^2 + 6)(p^2 - 4) \ge \left(x + \frac{19}{8x}\right)^4$$ or $$\frac{4992x^4 + 126112x^2 - 130321}{4096x^4} \ge 0$$ which is true.

We are done.

Answer 2

It's wrong. Try $x\rightarrow0$ and $\sin{y}\neq0$.

About your second problem.

We need to prove that: $$|\sin{y}|\left(x+\frac{1}{x}-2\cos{y}\right)\leq\sqrt{x^2+6}$$ and it's enough to prove that $$\sin{y}\left(x+\frac{1}{x}+2\cos{y}\right)\leq\sqrt{x^2+6},$$ where $y\in\left[0,\frac{\pi}{2}\right].$

I hope it will help.

Answer 3

Fix $x$ and put $X=x+\frac 1x$. Note that $X^2\le x^2+3$. Put $f(y)=\sin{y}\cdot(X+2\cos{y})$. As Michael Rozenberg noted in his answer, it is enough to prove that $f(y)\le \sqrt{x^2+6}$ for each $y\in\left[0,\frac{\pi}{2}\right]$. Since the function $f$ is continuous and the set $\left[0,\frac{\pi}{2}\right]$ is compact, the function $f(y)$ attains its maximum on $\left[0,\frac{\pi}{2}\right]$ at some its point $y_0$. The following cases are possible:

1)) $y_0=0$. Then $f(y_0)=0\le \sqrt{x^2+6}$.

2)) $y_0=\frac{\pi}{2}$. Then $f(y_0)=X\le \sqrt{x^2+3}<\sqrt{x^2+6}$.

3)) $0=f'(y_0)=X\cos y_0+4\cos^2 y_0-2$. Put $Y=\cos y_0$. The roots of the quadratic equation for $Y$ are $\frac{-X\pm \sqrt{X^2+32}}{8}$. Since $X, Y\ge 0$, we have $Y=\frac{\sqrt{X^2+32}-X}{8}$. Then $$f(y_0)^2=(1-Y^2)(X+2Y)^2=$$ $$\frac 1{128}\left(X^3\sqrt{X^2+32}-X^4+80X^2+32X\sqrt{X^2+32}+128\right).$$

But $(\sqrt{X^2+32}-X)(\sqrt{X^2+32}+X)=X^2+32-X^2=32$, so $$\sqrt{X^2+32}-X<\frac {32}{2X}.$$

Thus $$128f(y_0)^2<X^3\frac{16}{X}+48X^2+32X\frac{16}X+128=$$ $$64X^2+640\le 64(x^2+13)\le 128(x^2+6).$$

Answer 4

By Alex's substitution $u=x+\frac1x>2$ $$I=\frac{\sin^2y\,(x+\frac1x-2\cos y)^2}{x^2+6}\leq\frac{\sin^2y(u-2\cos y)^2}{u^2+3}:=f(u)$$ Then $f(u)$ has a critical point at $u=-\frac3{2\cos y}$ for any fixed $y$ with $I=\sin^2y+\frac13\sin^22y\leq\frac{103}{100}.$ Unfortunately it exceeds the bound 1 a little. One must modify $f(u)$.

Also, $$ f(2)=\frac{16\sin^2y\sin^4(\frac y2)}{7}\leq\frac{27}{28}.$$

Edit:

I tried without substitution. The critical point of $I=I(x)$ for a fixed $y$ is the solution of the cubic $\cos y\,x^3+2x^2-3=0$. Then by substituting $\cos y=\frac{3-2x^2}{x^3}$ in $I(x,y)$ we have $$I(x)=\frac{(x^6-4x^4+12x^2-9)(x^2+6)(x^2-1)^2}{x^{12}}$$ which has the local maximum $\frac{16807}{18432}$ at $x=2\sqrt{\frac35}$.