Show that- $$x-\frac{x^2}{2}&lt;\ln(1+x)&lt;x-\frac{x^2}{2(1+x)}, x&gt;0$$  I can prove this by observing that at $x=0$, the three functions in the above inequality (say $f_1&lt;f_2&lt;f_3$) are zero. And for $x&gt;0$, the derivative of the difference of $f'_2-f'_1&gt;0$ and $f'_3-f'_2&gt;0$ therefore the inequality is satisfied.   I am looking for an elegant solution to this using some kind of Mean Value Theorem maybe, if possible.
121537
To prove: $$x-\frac{x^2}{2}&lt;\ln(1+x)&lt;x-\frac{x^2}{2(1+x)},\quad\forall x&gt;0$$  I have used Taylor series expansion at 0 for both the inequalites. The greater than by expanding $\ln(1+x)$ and the less than by expanding $\int \ln(1+x)\,dx$ at 0.   Is there a cleaner / more elegant way of achieving the same?
76016
We know that for all $\varepsilon &gt; 0$, there exists an $N \in \mathbb{N}$ such that $\lvert x_n - 2 \rvert &lt; \epsilon$ for all $n \geq N$, and we want to show that for all $\varepsilon' &gt; 0$, there exists an $N' \in \mathbb{N}$ such that $\left| \frac{1}{x_n} - \frac{1}{2} \right| &lt; \epsilon'$ for all $n \geq N'$.  Let $\varepsilon = \varepsilon' - \frac{3}{2}$. Then by the triangle inequality, we have $$\left| \frac{1}{x_n} - \frac{1}{2} \right| \leq \left| \frac{1}{x_n}\right| + \left| -\frac{1}{2}\right|.$$  Because we proved earlier that $x_n &gt; 1$ for all $n \geq N$, we know that $$\left| \frac{1}{x_n}\right| + \left| -\frac{1}{2}\right| &lt; \left| x_n\right| + \left| -\frac{1}{2}\right|.$$  By the triangle inequality again, we know that $$\left| x_n - \frac{1}{2}\right| \leq \left| x_n\right| + \left| -\frac{1}{2}\right|.$$  Note that $\left| x_n - \frac{1}{2}\right| = \left| (x_n - 2) + (2 - \frac{1}{2})\right|$, so we get $$\left| (x_n - 2) + (2 - \frac{1}{2})\right| \leq \left| x_n - 2\right| + \left| 2 - \frac{1}{2}\right|.$$  Then we have $$\left| x_n - 2\right| + \left|2 - \frac{1}{2}\right| &lt; \varepsilon + \frac{3}{2} = \left(\varepsilon' - \frac{3}{2}\right) + \frac{3}{2} = \varepsilon',$$ so $\frac{1}{x_n} \rightarrow \frac{1}{2}$.  I just realized that I don't know how to make sure that $\varepsilon' - \frac{3}{2}$ is a positive number. How can I do this?
97847
Suppose that g is a function that is continuous on $[0,1]$ and differentiable on $(0,1)$, $g(0)=0$ and $g(1)=1$. Show that:  1) There exists $a_1&lt;a_2$ in $(0,1)$ such that $g'(a_1)+g'(a_2)=2$   2) There exists $a_1&lt;a_2$ in $(0,1)$ such that $\frac{1}{g'(a_1)}+ \frac{1}{g'(a_2)}=2$   3) There exists $a_1&lt;a_2$ in $(0,1)$ such that $g'(a_1)g'(a_2)=1$   I know by the Mean Value Theorem, there exists $a$ such that $f'(a)=1$, but I am not sure on how am I supposed to show the existence of two items satisfying the conditions.  Many thanks for your help on this!
49841
Let $x_1=a&gt;0$ and $x_{n+1}=x_n+\frac{1}{x_n} \forall n\in \mathbb N$.   Check whether the following sequence converges or diverges.   When I was in UG my teacher used derivative test for monotonicity.  $f(x)=x+\frac{1}{x}, f'(x)=1-\frac{1}{x^2}&gt;0(x&gt;1).$ So, $f(x)$ is increasing. How to prove the sequence is monotonic? Differentiation is coming after the sequences and series. By AM-GM inequality sequence is bounded below. $x_{n+1}=x_n+\frac{1}{x_n}\ge 2\sqrt{x_n.\frac{1}{x_n}}=2  \forall n\in \mathbb N$. How can I judge whether the sequence bounded above or not? Please help me.
122170
Suppose $f(0) =0 $ and  $0&lt;f''(x)&lt;\infty (\forall$  $x&gt;0)$, then $\frac{f(x)}{x}$ strictly increases as $x$ increases.    I have shown that $f'(x)-\frac{f(x)}{x}  = \frac{1}{2}xf''(c)$, for some $c\in (0,x)$. How do I proceed from here?
106250
Given Question is Prove that  $$\frac{b-a}{\sqrt{1-a^2}}&lt;\sin^{-1}b-\sin^{-1}a&lt;\frac{b-a}{\sqrt{1-b^2}}$$ when $0&lt;a&lt;b&lt;1$ using Mean Value Thoerem.  I considered $f(x)=\sin^{-1}x $ when $x\in(0,b)$ and used LMVT and established that $\frac{b}{\sqrt{a-c^2}}=\sin^{-1}b$ where $c \in (0,b)$.  But I don't know how to proceed after this. Please guide.
116708
Prove: $$\frac{1}{\sqrt{x}}\ge \frac{2}{x+1}, \quad\forall x&gt;0$$  Yeah, pretty much it. I've tried all manner of rearranging and just can't seem to get it. Thanks.
31460
Let $f:\mathbb{R} \to \mathbb{R}$ be a twice differentiable function. And let  $$\eqalign{   &amp; M_0  = \sup \left|f(x)\right|  \cr                &amp; M_1  = \sup \left|\frac{d}{dx} f(x) \right|  \cr                &amp; M_2  = \sup \left|\frac{d^2}{dx^2} f(x) \right| \cr }$$ Prove that  $$M_1 ^2  \leqslant 4M_0 M_2$$ I can not think how I can relate these values ​​in some inequality )=
101338
How can I prove that $$\displaystyle x-\frac {x^2} 2 &lt; \ln(1+x)$$ for any $x&gt;0$  I think it's somehow related to Taylor expansion of natural logarithm, when:  $$\displaystyle \ln(1+x)=\color{red}{x-\frac {x^2}2}+\frac {x^3}3 -\cdots$$  Can you please show me how? Thanks.
99631
I have tried the following: $$ f(x) = \sin x-\frac{2x}{\pi} \\ f'(x)= \cos x-\frac{2}{\pi} \\ f''(x) = -\sin x \leq 0 $$ But this doesn't seem to be heading in the right direction as it would appear $f'(x)$ is decreasing and has a zero within the domain, so I'm not sure how this might be used to prove the inequality.
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?
24299
Suppose $x \in (0,\frac{\pi}{2})$ and we want to prove $$x&lt;\frac{\sin x +\tan x}{2}$$I tried to prove it by taking $f(x)=\sin x+ \tan x -2x$ and show $f(x) &gt;0 ,when\space  x \in (0,\frac{\pi}{2})$ take f'$$f'=\cos x +1+\tan ^2 x-2\\=\tan^2 x-(1-\cos x)\\=\tan ^2 x-2sin^2(\frac x2)$$ I get stuck here ,because the last line need to be proved $\tan ^2 x&gt;2sin^2(\frac x2) ,when\space  x \in (0,\frac{\pi}{2})$ $\bf{Question}:$ Is there a geometrical method to prove the first inequality ? (or other idea)  Thanks in advance.    $\bf{Remark}: $I can see the function is increasing $when\space  x \in (0,\frac{\pi}{2})$ like below :
115973
I'm not sure how to approach this question, but I know it has something to do with taking the derivative of this. Can anyone please help me out?  Let x > 0. Prove that the value of the following expression does not depend on x: $$\int_{0}^{x}\frac{1}{1+t^4} dt + \frac{1}{3}\int_{0}^\frac{1}{x^3} \frac{1}{1+t^\frac{4}{3}} dt$$
92685
Can you help me with this inequality?  $|x-1| +|x+1| &lt;1$.  I think that it doesn't exist an $x$ that satisfies this inequality, but I'm not really convinced about my answer.
16150
How prove  this inequality $$\tan{(\sin{x})}&gt;\sin{(\tan{x})},0&lt;x&lt;\dfrac{\pi}{2}$$  this  PDF give  a ugly methods :   I post this solution:let     $$f(x)=\tan{(\sin{x})}-\sin{(\tan{x})}$$   then we have   $$f'(x)=sec^2{(\sin{x})}\cos{x}-\cos{(\tan{x})}\sec^2{x}=\dfrac{\cos^3{x}- \cos{(\tan{x})}\cos^2{(\sin{x})}}{\cos^2{(\sin{x})}\cos^2{x}}$$   case 1:$0&lt;x&lt;\arctan{\dfrac{\pi}{2}}$, then we have    $$0&lt;\tan{x}&lt;\dfrac{\pi}{2},0&lt;\sin{x}&lt;\dfrac{\pi}{2}$$   so Use AM-GM inequality we have   $$\sqrt[3]{\cos{(\tan{x})}\cos^2{(\sin{x})}}\le\dfrac{1}{3}[\cos{(\tan{x})}+2\cos{(\sin{x})}]\le\cos{\dfrac{\tan{x}+2\sin{x}}{3}}$$   use   $$\tan{x}+2\sin{x}&gt;3x$$   so   $$f'(x)&gt;0\Longrightarrow f(x)&gt;0$$   case2: $\arctan{\dfrac{\pi}{2}}\le x\le\dfrac{\pi}{2}$, so   $$\sin{(\arctan{\dfrac{\pi}{2}})}&lt;\sin{x}&lt;1$$   since   $$\sin{(\arctan{\dfrac{\pi}{2}})}=\dfrac{\pi}{\sqrt{4+\pi^2}}&gt;\dfrac{\pi}{4}$$   $$\Longrightarrow \dfrac{\pi}{4}&lt;\sin{x}&lt;1$$,so   $$1&lt;\tan{(\sin{x})}&lt;\tan{1},$$,so   $$f(x)&gt;0$$   I think this inequality have other  simple methods.Thank you  It is said can use integral inequality,But I can't   In fact,we have $$\tan{(\sin{x})}-\sin{(\tan{x})}=\dfrac{1}{30}x^7+o(x^7),x\to 0$$ can see:
121784
I looking for help with proving the following inequality. Any relevant logarithmic identities would be great. Tried differentiating and taking limits and I'm lost as to how to approach this.   $$\frac 1{x+1}&lt;\ln\left(\frac{x+1}x\right)&lt;0.5\left(\frac 1x+\frac 1{x+1}\right)$$
63104
Let $a&gt;b&gt;0$ and let $n \in N$ satisfy $n \geq 2$.  Prove that $$a^{1/n}-b^{1/n}&lt;(a-b)^{1/n}.$$  If we let $a=x$ and $b=x-1$, then we need to show that $f(x):=x^{1/n}-(x-1)^{1/n}$ is decreasing for $1\leq x$ (which means that $f'(x)\leq 0$ and $f$ is differentiable), and evaluate $f$ at $1$ and $a/b$.