How to prove $1+(1/2)x-(1/8)x^2&lt;\sqrt{1+x}$ for $x&gt;0$ by Taylor Expansions up to $n$ order or Mean Value Theorem? I tried to apply MVT on $\sqrt{1+x}$ and get $\displaystyle \sqrt{1+x}=1+\frac{1}{2\sqrt{1+\xi}}x$ for $\xi\in (0,x)$. How to do next?
122721
I need to show that for $x&gt;0$, $1+\frac{x}{2} \ge \sqrt{1+x} \ge 1+\frac{x}{2} - \frac{x^2}{8} $.  I used the geometric/arithmetic mean inequality to show that $1+\frac{x}{2} \ge \sqrt{1+x}$ is indeed true. My issue lies in the second part of this.  In trying to show that $\sqrt{1+x} \ge 1+\frac{x}{2} - \frac{x^2}{8} $, I determined that this is a partial taylor series expansion and I attempted to play around with both inequalities to show that this is true but to no avail.  I would appreciate any help in solving this. Ideally I would like to do this analytically but any solutions using more complicated theorems are welcome.
75997
How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?   im trying to use AM-GM  $\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}}   \le \sum_{k=1}^n \frac{{\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}}{n}$  $ln\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}} \le ln \frac{1}{n} \sum_{k=1}^n {\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}$  $ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}\le  \sum_{k=1}^n {\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}$  im not sure is this right or not, however i dont know how to include the $n^2$ in the nominator? is there also alternative proof using jensen inequality?
110245
Problem.  Show that $$e^x &gt; 1 + x , \ \ x \neq 0$$  My attempt.  Using Mean Value Theorem: $$f'(c) = \frac{f(b) - f(a)}{b-a}$$  $$\Rightarrow e^c = \frac{e^x - 1}{x}$$ $$\Rightarrow xe^c = e^x -1$$ $$\Rightarrow x+1 = \frac{e^x}{e^c}$$  I'm not feeling very comfortable with these calculations...Am I on the right track? Can I even choose $e^o$ in my initial calculation as it says that $x \neq 0 $ ?
70462
I want to show that $$\frac{x}{1+x}&lt;\log(1+x)&lt;x$$ for all $x&gt;0$ using the mean value theorem. I tried to prove the two inequalities separately.  $$\frac{x}{1+x}&lt;\log(1+x) \Leftrightarrow \frac{x}{1+x} -\log(1+x) &lt;0$$  Let $$f(x) = \frac{x}{1+x} -\log(1+x).$$ Since $$f(0)=0$$ and $$f'(x)= \frac{1}{(1+x)^2}-\frac{1}{1+x}&lt;0$$ for all $x &gt; 0$,  $f(x)&lt;0$ for all $x&gt;0$.  Is this correct so far?  I go on with the second part: Let $f(x) = \log(x+1)$. Choose $a=0$ and $x&gt;0$ so that there is, according to the mean value theorem, an $x_0$ between $a$ and $x$ with   $f'(x_0)=\frac{f(x)-f(a)}{x-a} \Leftrightarrow \frac{1}{x_0+1}=\frac{ \log(x+1)}{x}$.  Since $$x_0&gt;0 \Rightarrow  \frac{1}{x_0+1}&lt;1.$$ $$\Rightarrow 1 &gt; \frac{1}{x_0+1}= \frac{ \log(x+1)}{x} \Rightarrow x&gt; \log(x+1)$$
48882
Towards the end of , how do I get  $$ \frac{\sqrt{8z + 1} - 1}{2} &lt; w + 1 $$  Here $w = x + y \geq 0, t = (w^2 + w)/2 , z = t + w, x,y \in \mathbb N_{\geq 0}$.  Thanks.
107470
Find the smallest value of $c$ such that $(1-x)\ln(1-x)+(1+x)\ln(1+x)\leq c x^2$ holds for $0&lt;x&lt;1$.   I saw the curve and realized this is true for $c=2$. How can I prove it? What is the smallest $c$ that still makes the inequality valid?  I think we should use the Taylor expansions of the $\ln$s.
105755
I think I have the first part, as I have shown that $q &lt; q + (q'-q)/2 &lt; q'$, but I have trouble in proving that $|q-q'|&lt; \epsilon$.  Could someone tell me if there is a better way of showing the first part and how to show the second part?  For those asking x is a real number
35318
I am trying to prove the existence of the square root of $2$.  I have some steps with a very vague explanation and I would like to clarify.  The proof: Let  $$S=\{x\in\mathbb R\mid x\geqslant 0 \text{ and } x^2&lt;2\}.$$  I understand the proof of LUB, ∝   and so I am at the step where $\alpha^2=2$.  I know that we are to prove by contradiction so we state let $\alpha^2 &lt;2$ and $\alpha^2 &gt;2$.  Now my instructor wants us to use the Archimedean Axiom $1/n = \varepsilon$.    $(\alpha^2 + 1/n)^2$  then what.....
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.
66170
Let $f$ be defined on $\mathbb{R}$ by $f(x) = e^{-1/x^2}$ for $x$ not equal to $0$. and $f(0)= 0$. Prove that $f^{(n)}(0)=0$ for all $n = 1, 2,3$ ...  Do I need to use Taylor expansion from calculus class?  Any hint would be appreciated.
16469
I have seen a proof that shows $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n = e^x$$ by looking at the Taylor series expansion of $\ln(1+x)$ at $x=0$. To prove a theorem, my textbook uses the fact $$\lim_{n \to \infty} \left(1+\frac{a_n}{n}\right)^n = e^a$$ when $a_n \to a$. How can I prove this?
88959
Prove that $$\frac{x^2}{1+x^4} \leq \frac{1}{2}.$$   First off, we observe that the expression on the LHS is positive for all $x \in \Bbb R,$ and equality is achieved iff $x \in \{-1, 1 \}$. That being said, we start by manipulating the expression as below;  \begin{align} \frac{x^2}{1+x^4} - \frac{1}{2} &amp;\leq 0\\\\ \impliedby\frac{2x^2 - (1+x^4)}{2(1+x^4)} &amp;\leq 0\\\\ \impliedby -\frac{(x^4 - 2x^2 + 1)}{2(1+x^4)}&amp;\leq 0\\\\ \impliedby \frac{x^4 - 2x^2 + 1}{2(1+x^4)} &amp;\geq 0\\\\ \end{align}   We then make the observation that $x^4 - 2x^2 + 1 = (x^2 - 1)^2 \geq 0$ for all $x \in \Bbb R$ and that $1 + x^4 \geq 1$ for all $x \in \Bbb R$, and so the final inequality holds for all $x \in \Bbb R. \\\hspace{92pt}\square$  Is it okay to prove it in this way? I have no solutions so it would be nice if I could have some validation!
72110
Prove that $\frac{1}{a+1}&lt;\ln \frac{a+1}{a}&lt;\frac{1}{a},a&gt;0$  First inequality using MVT:  $\frac{1}{a+1}&lt;\ln \frac{a+1}{a}:$  $f(a)=\frac{1}{a+1}-\ln \frac{a+1}{a}$  $f(1)=\frac{1-2\ln 2}{2},f^{'}(a)=\frac{1}{a(a+1)^2}&gt;f(1)\Rightarrow f(a)&gt;f(1)$  $\frac{1}{a+1}-\ln \frac{a+1}{a}-\frac{1-2\ln 2}{2}&gt;0$  This is not the starting inequality.   Is there something wrong in this method?
25334
I have a math homework where it's being asked to prove that :  $$\forall a \geq 0,\sqrt{a}\leq\frac{1+a}{2}$$  However, I don't have any idea how I should start this one...  Any idea ?
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$$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt n}\geq \sqrt n$$ I want to prove this by Induction  $$n=1 \checkmark\\ n=k \to \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt k}\geq \sqrt k\\ n=k+1 \to \dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt {k+1}}\geq \sqrt {k+1}$$ so $$\dfrac{1}{\sqrt 1}+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+\cdots+\dfrac{1}{\sqrt k}+\dfrac{1}{\sqrt {k+1}}\geq \sqrt k+\dfrac{1}{\sqrt {k+1}}$$now we prove that $$\sqrt k+\dfrac{1}{\sqrt {k+1}} &gt;\sqrt{k+1} \\\sqrt{k(k+1)}+1 \geq k+1 \\ k(k+1) \geq k^2 \\k+1 \geq k \checkmark$$ and the second method like below ,   and I want to know is there more Idia to show this proof  ? forexample  combinatorics proofs  , or using integrals ,or fourier series ,....    Is there a close form for this summation ?  any help will be appreciated .
29500
Prove that for any positive integer $n$,       $$n^{1/n}  &lt; 1 + \sqrt{\frac{2}{n}}.$$   This due to Victor Linis, Eureka, Vol. 2, No. 2, February 1976, p. 29.   Hint:      Use the binomial theorem.
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In an ordered field show that $x \geq 0 \implies (1+x)^{n} \geq 1+nx+ \frac{1}{2}n(n-1)x^2$ for every positive integer $n$.   I know that $(1+x)^{n} \geq 1+nx$ (Bernoulli's inequality). To get the stronger inequality you can probably use induction also. But is there an easier way? The extra term on the right seems to be an "error" term and looks like a Taylor series expansion.