# Problem 790 from Pentagon: an Inequality in Two Variables

### Solution 1

\displaystyle \begin{align} \cos \Bigr(2x+\frac{\pi}{6}\Bigr)\leq 1&\Leftrightarrow \frac{\sqrt{3}}{2}\cos 2x-\frac{1}{2}\sin 2x\leq 1\\ &\Rightarrow \int_a^b \frac{2 \cos 2x}{2+\sin 2x}dx\leq \int_a^b \frac{2\sqrt{3}dx}{x}\\ &\Leftrightarrow [\ln |2+\sin 2x|]_a^b\leq \frac{2\sqrt{3}}{3}(b-a)\\ &\Leftrightarrow \ln \Bigr|\frac{2+\sin 2b}{2+\sin 2a}\Bigr|\leq \frac{2\sqrt{3}}{3}(b-a) \end{align}

### Solution 2

Let $f(x)=\ln (2+\sin 2x), x\in \mathbb{R}.$ Then

\displaystyle \begin{align} f'(x)&=\frac{2\cos 2x}{2+\sin 2x}\\ f''(x)&=2\frac{(-2\sin 2x)(2+\sin 2x)-(2\cos 2x)\cos 2x}{(2+\sin 2x)^2}\\ &=(-4)\cdot \frac{2 \sin 2x+1}{(2+\sin 2x)^2} \end{align}

\displaystyle \begin{align} f''(x)&=0\Leftrightarrow \sin 2x=-\frac{1}{2}=\sin \Bigr(\frac{7\pi}{6}\Bigr)\\ &\Leftrightarrow 2x=n\pi +(-1)^n \Bigr(\frac{7\pi}{6}\Bigr), n\in \mathbb{Z}\\ &\Leftrightarrow x=n\Bigr(\frac{\pi}{2}\Bigr)+(-1)^{n+1}\frac{\pi}{12}, n\in \mathbb{Z} \end{align}

Thus,

\displaystyle\begin{align} \max f'(x)&=\max \left\{f'\left(\frac{7\pi}{12}\right), f'\left(\frac{11\pi}{12}\right),f' \left(\frac{19\pi}{12}\right),\ldots \right\}\\ &=\max \left\{f'\left(\frac{7\pi}{12}\right), f'\left(\frac{11\pi}{12}\right)\right\}\\ &=\frac{2}{\sqrt{3}}\\ \min f' (x)&=-\frac{2}{\sqrt{3}}. \end{align}

By the first mean value theorem, for $a\lt b,$

\displaystyle \begin{align} \frac{f(b)-f(a)}{b-a}=f'(c),\, \textit{ for some }\,c\in (a,b)&\Rightarrow\\ \ln \Bigr(\frac{2+\sin 2b}{2+\sin 2a}\Bigr)=f'(c)(b-a)&\Rightarrow\\ \Bigr|\ln \Bigr(\frac{2+\sin 2b}{2+\sin 2a}\Bigr)\Bigr|=|f'(c)|(b-a)\leq \frac{2}{\sqrt{3}}(b-a). \end{align}

### Solution 3

Let $f(x)=\ln \left(2+\sin (2x)\right).$ Clearly, $f$ is differentiable on $\mathbb{R}$ and we have

\displaystyle \begin{align} f'(x)&=\frac{2\cos (2x)}{2+\sin (2x)}\\ f''(x)&=-4\cdot \frac{1+2\sin (2x)}{(\sin (2x)+2)^2} \end{align}

Now, $f''$ vanishes at $\displaystyle x_0=-\frac{\pi}{12} (\mod \pi).$ Hence, $f'$ has a maximum at $\displaystyle x_0=-\frac{\pi}{12}$ and we have $\forall x\in \mathbb{R},$ $\displaystyle |f'(x)|\leq \frac{2\sqrt{3}}{3}.$

Using the Lagrange's theorem, there is $c\in (a,b)\quad [\textit{for sure } a,b\in \mathbb{R}]$ such that $\displaystyle \frac{f(b)-f(a)}{b-a}=f'(c)$ hence we conclude that $\displaystyle |f(b)-f(a)|\leq \frac{2\sqrt{3}}{3}|f'(c)|$ or $\displaystyle \left|\ln \left(\frac{2+\sin (2b)}{2+\sin (2a)}\right)\right|\leq \frac{2\sqrt{3}}{3}|b-a|$ and we are done.

### Solution 4

Let $f(x)=\ln |2+\sin 2x|,$ $\displaystyle f'(x)=\frac{2\cos 2x}{2+\sin 2x}.$ By Lagrange's theorem, $f(b)-f(a)=f'(c)(b-a),$ with $c\in (a,b).$ Specifically,

$\displaystyle \ln \left|\frac{2+\sin 2b}{2+\sin 2a}\right|\leq \frac{2\cos 2c}{2+\sin 2c}(b-a)$

We need to prove that

\displaystyle \begin{align} \frac{2\cos 2c}{2+\sin 2c}\leq \frac{2\sqrt{3}}{3}&\Leftrightarrow \sqrt{3}\cos 2c-\sin 2c\leq 2\\ &\Leftrightarrow \cos \left(2c+\frac{\pi}{6}\right)\leq 1. \end{align}

True.

### Solution 5

The given inequality transforms into:

\displaystyle \begin{align} &\ln |2+\sin 2b|-\ln |2+\sin 2a|\leq \frac{2\sqrt{3}}{3}(b-a)\\ &\Leftrightarrow \ln (2+\sin 2b)-\ln (2+\sin 2a) \overbrace{\leq}^{(a)}\frac{2\sqrt{3}}{3}(b-a) \end{align}

(for $2+\sin 2t\geq 1\gt 0\;\forall t\in \mathbb{R},$ as $-1\leq \sin 2t \leq 1).$

Let $f(x)=\ln (2+\sin 2x).$ Then $\displaystyle (a) \Leftrightarrow f(b)-f(a)\leq \frac{2\sqrt{3}}{3}(b-a).$

Clearly, $f(x)$ is continuous and differentiable on any interval $(a,b).$ Therefore, using Lagrange's Mean Value Theorem, there exists a value $c \ in (a,b),$ such that $\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}.$ Now, with

$\displaystyle g(x)=f'(x)=\frac{2\cos (2x)}{\sin 2x+2},$ $\displaystyle g'(x)=-\frac{4(1+2\sin 2x)}{(\sin 2x+2)^2},$ and $\displaystyle g''(x)=\frac{(16\cos 2x)(\sin 2x-1)}{(\sin 2x+2)^3}.$

Now, $g'(x)=0$ when $\displaystyle \sin 2x=-\frac{1}{2}$ and $\displaystyle \cos 2x=\pm \frac{\sqrt{3}}{2}.$

When $\displaystyle \sin 2x=-\frac{1}{2}$ and $\displaystyle \cos 2x=\frac{\sqrt{3}}{2},$ $g''(x)\lt 0,$ implying $f(x)$ attains maxima, and, when $\displaystyle \sin 2x =-\frac{1}{2}$ and $\displaystyle \cos 2x=-\frac{\sqrt{3}}{2},$ $g''(x)\gt 0,$ implying $f(x)$ attains minima. Therefore

$\displaystyle g(x)_{\max} =\frac{2(\frac{\sqrt{3}}{2})}{2-\frac{1}{2}}=\frac{\sqrt{3}}{\frac{3}{2}}\overbrace{=}^{(3)} \frac{2\sqrt{3}}{3}$

From (1), (2), (3), it follows that

$\displaystyle\frac{f(b)-f(a)}{b-a}\leq \frac{2\sqrt{3}}{3}\Rightarrow f(b)-f(a)\leq \frac{2\sqrt{3}}{3}(b-a),$

because $b-a\gt 0.$

### Solution 6

Let $\displaystyle \angle POx=\frac{\pi}{2}-2x$ for $\displaystyle P\left(\cos \left(\frac{\pi}{2}-2x\right)\right).$ We see that the slope of $AP$ is less than that of $AT.$ So that

$\displaystyle \frac{\sin (\frac{\pi}{2}-2x)}{\cos (\frac{\pi}{2}-2x)+2}\leq \frac{1}{\sqrt{3}}.$

Therefore

$\displaystyle \int_a^b \frac{2\cos 2x}{2+\sin 2x}dx\leq \int_a^b \frac{2}{a\sqrt{3}}dx\Leftrightarrow [\ln |2+\sin x|]_a^b \leq \frac{2}{\sqrt{3}}(b-a)$

### Solution 7

Condition $a\lt b$ is unnecossary. Let $f(x)=\ln(2+\sin 2x).$ Then $\displaystyle |f(b)-f(a)=\frac{2|\cos 2c|}{2+\sin 2c}\cdot |b-a|.$ Hence, suffice it to show that $\displaystyle\frac{2|\cos 2c|}{2+\sin 2c}\le\frac{2}{\sqrt{3}}$ which is equivalent to $-\sin 2c\pm\sqrt{3}\cos 2c\le 2.$ That can be verified by squaring or with the Cauchy-Schwarz inequality.\$

### Acknowledgment

I could not be more grateful to Dan Sitaru who has kindly communicated to me the problem with several solutions on a LaTeX file. Solution 1 is by Abdelhak Maoukouf (Casablanca, Morocco); Solution 2 by Ravi Prakash (New Delhi, India); Solution 3 by Anas Adlany (El Zemamra, Morocco); Solution 4 by Richdad Phuc (Hanoi, Vietnam); Solution 5 by Soumava Chakraborty (Kolkata, India); Solution 6 by Kunihiko Chikaya (Tokyo, Japan); Solution 7 by Leo Giugiuc (Drobeta-Turnu Severin, Mehedinti, Romania).

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