# How prove $\bigl(\frac{\sin x}{ x}\bigr)^{2} + \frac{\tan x }{ x} >2$ for $0 < x < \frac{\pi}{2}$

How prove $\left(\frac{\sin x}{ x}\right)^{2} + \frac{\tan x }{ x} >2$ for $0 < x < \frac{\pi}{2}$. Can this be proved with simple way?

$$\left(\frac{\sin x}{x}\right)^2 + \left(\frac{\tan x}{x}\right) = \\ \left(\frac{x - \frac{x^3}{6} + \frac{x^5}{120} + \ldots}{x}\right)^2 + \left(\frac{x + \frac{x^3}{3} + \frac{2x^5}{15} + \ldots}{x}\right) = \\ \left(1 - \frac{x^2}{6} + \frac{x^4}{120} - \ldots\right)^2 + \left(1 + \frac{x^2}{3} + \frac{2x^4}{15} + \ldots\right) \simeq \\ \left(1 + 2 \left(-\frac{x^2}{6} + \frac{x^4}{120} - \ldots \right)\right) + 1 + \frac{x^2}{3} + \frac{2x^4}{15} + \ldots\\2 + x^2 \left(\frac{ - 2}{6} + \frac{1}{3}\right) + x^4 \left(\frac{2}{120} + \frac{2}{15} \right) + \ldots = \\2 + x^4 \left(\frac{9}{60} \right) + \ldots > 2.$$

• Could you please explain a bit more how this gives a proof? I'm a bit worried about the $\ldots$. Are you able to prove that all terms in the expansion comes with the right sign? Since people are voting your answer up, it is probably me being slow... – mickep Dec 30 '14 at 19:08
• I am not so sure about the use of the asymptotic equivalence over a given interval, rather than an interval which we may choose to be arbitrarily small (but nondegenerate). I agree that this definitely works over a small interval near $0$, however. – Ian Dec 30 '14 at 19:10
• try it by 3 term , expand it for sure – Khosrotash Dec 30 '14 at 20:15
• I agree that the terms seem to be positive (but I have no proof). I get that the term in front of $x^4$ is $8/45$ and the term in front of $x^6$ is $16/315$. But this doesn't prove anything (yet). – mickep Dec 30 '14 at 20:32
• This is not a proof. You have not shown that you can ignore the omitted terms for all x. – marty cohen Dec 8 '16 at 23:19

Set $y = 2\log\left(\frac{x}{\sin x}\right)$ and $z=\log\left(\frac{\tan x}{x}\right)$.

Exploting Weierstrass products, for every $x\in I=\left(0,\frac{\pi}{2}\right)$ we have: $$z-y = \sum_{n=1}^{+\infty}\sum_{m=1}^{+\infty}\frac{x^{2m}}{m\cdot \pi^{2m}}\left(\frac{4^m}{(2n-1)^{2m}}-\frac{3}{n^{2m}}\right),$$ so, given that $\zeta(2m) = \sum_{n=1}^{+\infty} \frac{1}{n^{2m}},$ $$z-y = \sum_{m=1}^{+\infty}\frac{x^{2m}\cdot\zeta(2m)}{m\cdot \pi^{2m}}(4^m-4).\tag{1}$$ Since every term in the RHS of $(1)$ is non-negative, it follows that $z > y$ for every $x\in I$.

So we have: $$\left(\frac{\sin x}{x}\right)^2+\frac{\tan x}{x} = e^z + e^{-y} > e^{y}+e^{-y} \geq 2. \tag{2}$$

One could look at the function and see that it is monotonously rising in the given interval, one would have to look a the limits, if $x$ approaches $0$ or $\pi/2$. This should work as well. Would you know, how to do that?

• The problem is that although the second term is increasing, the first term is decreasing. Showing that the second term is increasing faster than the first is decreasing doesn't seem obvious to me (and taking the derivative doesn't immediately expose anything useful to a proof). – Ian Dec 30 '14 at 17:30
• I agree. This way might also be not as interesting or elegant as the one's already stated here, especially the one by darya is nice. – Imago Dec 30 '14 at 18:31

this is famous GENERALIZED WILKER inequality : see this interesting some reslut with two papers 1 and 2.

since $\sin(x)>0$ and $x>0$ and $\tan(x)>0$ we have $$\left(\frac{\sin(x)}{x}\right)^2+\frac{\tan(x)}{x}\geq 2\sqrt{\left(\frac{\sin(x)}{x}\right)^2\cdot \frac{\sin(x)}{x}\cdot\frac{1}{\cos(x)}}$$ Now we have to show that $$\frac{\sin(x)^3}{x^3}>\cos(x)$$ in the given interval, defining $$f(x)=\frac{\sin(x)^3}{x^3}-\cos(x)$$ and we get $$\lim_{x \to 0+}f(x)=0$$ and $$f'(x)=-{\frac {\sin \left( x \right) \left( -{x}^{4}-3\,\sin \left( x \right) \cos \left( x \right) x+3\, \left( \sin \left( x \right) \right) ^{2} \right) }{{x}^{4}}}$$ this is positive since $-\sin(x)<0$ and $$-x^4+3\sin(x)(\sin(x)-x\cos(x))<0$$.

• It might be worth commenting that the first inequality is a variant of the arithmetic-geometric mean inequality. It is definitely worth clarifying why the last inequality holds. – Ian Dec 30 '14 at 19:08
• the last inequality can also be shown with the method of calculus, namely the monotony – Dr. Sonnhard Graubner Dec 30 '14 at 19:22
• The last inequality is worth commenting, especially since $\sin x>x\cos x$ for $0<x<\pi/2$. – mickep Dec 30 '14 at 19:54