What is the proof for $\forall x \in [0, \frac{\pi}{2}]\quad \sin(x) \ge \frac{x}{2}$ ? Assuming it is true.
46536
I need to show that $\dfrac{2}{\pi}&lt;\dfrac{\sin(x)}{x}&lt;1$ for $0&lt;x&lt;\dfrac{\pi}{2}$.  I know I need to use the mean value theorem, would I just say that since $f$ is continuous in the interval (call it I).  We know $\exists c\in I$ such that $f'(c) = \dfrac{f\left(\dfrac{\pi}{2}\right)-f(0)}{\dfrac{\pi}{2}}$. I don't know where this is going though.  Also, would I need to show $\lim_{x \to 0}\dfrac{\sin(x)}{x} = 1$? Is there a way to do this without L'H Rule?
74372
Prove $\sin(x)\tan(x) &gt; x^2$ for $x \in ( \,0, \frac{\pi}{2}) \,$  So I did the following: Let $f(x) = \sin(x)\tan(x) - x^2$.  Then of course $f(0) = 0$ and I want to show that for $x \in ( \,0, \frac{\pi}{2}) \,$ $f'(x) &gt; 0$ but I don't really know where to go from here as I can't get anything reasonable done with the derivative.  I was thinking of approximating it using the following inequality $\sin(x) &lt; x &lt; \tan(x)$.   Is that a good direction?
84414
How to prove              $${{a}^{a}}{{b}^{b}}\ge {{\left(\frac{a+b}{2}\right)}^{a+b}}$$ $a&gt;0$,$b&gt;0$, thanks.
122986
Quick question, might end up having a simple answer, but I have here a "proof" that any integral from 0 to $2\pi$ is zero, as follows:  $$\int^{2\pi}_0f(x)dx$$  Now using u-substitution, let $u = \sin x$, so $dx = \frac{du}{\cos x}$, so:  $$\int^{0}_{0}\frac{f(x)du}{\cos x}$$  (The bounds were evaluated to zero due to the u-substituion).  However, this is zero, since the integral goes from zero to zero.  Any help as to why this is wrong would be appreciated.
41697
Prove that there exists a constant $c&gt;0$ such that for all $x \in [1,\infty)$, $$\sum_{n&gt;x} \frac{1}{n^{2}} \leq \frac{c}{x}$$
101806
If we assume that: $0\le x_1,x_2,\ldots,x_{10}\le\frac{\pi}{2} $ such that:  $$\sin^2 (x_1) +\sin^2 (x_2)+\cdots+\sin^2(x_{10})=1$$  How to prove that:  $$\frac{\cos(x_1) +\cos(x_2) +\cdots+\cos(x_{10})}{\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10})} \ge 3$$
66868
$X \sim \mathcal{N}(0,1)$, then to show that for $x &gt; 0$, $$ \mathbb{P}(X&gt;x) \leq \frac{\exp(-x^2/2)}{x \sqrt{2 \pi}} \&gt;. $$
81392
I want to show that for all $n\ge 2$, it holds that $$ \int \cos^n x\ dx = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n}\int \cos^{n-2} x\ dx. $$ I'm not even getting the result for the induction base $(n=2)$: Using integration by parts, I only get $$ \int \cos^2 x\ dx = \cos x \sin x + \int \sin^2 x\ dx. $$ I'm suspecting that I need to use some trigonometric identity here.
88631
Problem: Show that $\int_{0}^{\infty}|\frac{ \sin x}{x}|=\infty $ .  Here is what i have tried :  I found out that for all $x&gt;0$ , $x-\frac{x^3}{6} &lt; \sin x &lt;x $  . So $1-\frac{x^2}{6}&lt;\frac{ \sin x}{x}&lt;1$ . So applying integration to both side from $0$ to $M$ we have $M-\frac{M^3}{18}&lt;\int_{0}^{M}|\frac{ \sin x}{x}|&lt; M$ . Now we send $M \rightarrow \infty$ and observe that left-hand side goes to infinity . Hence proved .  Please point out if there is anything wrong with my approach . I will also be delighted if could provide alternative solutions . Thank you .
31979
Let $f(\sin x)$ be a given function of $\sin x$.   How would I show that $\int_0^\pi xf(\sin x)dx=\frac{1}{2}\pi\int_0^\pi f(\sin x)dx$?
112289
If $x \geq 2$, then how do we prove that $$\int_{2}^{x} \frac{\mathrm dt}{\log^{n}{t}} = O\Bigl(\frac{x}{\log^{n}{x}}\Bigr)?$$
88441
I would like to show that show $|t|\leq \pi$ implies $ |e^{it}-1|\geq 2|t|/ \pi$. I tried to proof it geometrically but without success.
87145
Do you see any nice way to prove that $$\arctan\frac{\pi}{2}\ge1 ?$$  Thanks!  Sis.
48695
I'm having some trouble figuring out where to start for this proof.     Prove for all positive integers $n$,   $$\sqrt{n+1}-\sqrt{n}\lt \frac{1}{2\sqrt{n}}$$   Any help will be greatly appreciated.
71323
Prove: $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$$ for all x, y positive   $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\geq 0$$  $$\frac{x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{x(\sqrt{x}-\sqrt{y})-y(-\sqrt{y}+\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(x-y)(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  All the elements are positive and if $\sqrt{x}=\sqrt{y}$ we get $0$  Is the proof valid?
120611
As my further preparation to Putnam competition, I came across such inequality to prove:  $$\forall n \in \mathbb{N}: \int_{0}^{\pi} \left|\frac{\sin(nx)}{x}\right|dx \geq \frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}$$  The problem is that I got stuck in spite of spending by far 6 days on this problem!  Because I got really stuck, I am very determined to see how to prove such inequality. Help very, very appreciated!
34073
How to prove the following product? $$\frac{\sin(x)}{x}= \left(1+\frac{x}{\pi}\right) \left(1-\frac{x}{\pi}\right) \left(1+\frac{x}{2\pi}\right) \left(1-\frac{x}{2\pi}\right) \left(1+\frac{x}{3\pi}\right) \left(1-\frac{x}{3\pi}\right)\cdots$$