$x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}&lt;\log(1+x)&lt;x-\frac{x^2}{2}+\frac{x^3}{3}$, $x&gt;0$  Here I can only see that the right side of second inequality i.e. $x-\frac{x^2}{2}+\frac{x^3}{3}$ comes in the expansion of $\log(1+x)$.  We have done the Lagrange's mean value theorem and intermediate value theorem, do these have anything to do with the inequality.  Kindly provide some hint.
120936
Prove that   $\displaystyle x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}&lt;\log(1+x) &lt;x-\frac{x^2}{2}+\frac{x^3}{3}$  My attempt: I can prove this by taking one side at a time and assuming $\displaystyle f(x) = \log (1+x)-x+\frac{x^2}{2}-\frac{x^3}{3}$   and then proving that it is a decreasing function and then the same for the other side.  But I am looking for a better solution using Lagrange's Mean Value Theorem or using Taylor's remainder.
68623
Is $\frac{\vert x+y\vert}{1+\vert x+y\vert}\leq\frac{\vert x\vert}{1+\vert x\vert}+\frac{\vert y\vert}{1+\vert y\vert}$ true for all $x,y\in\mathbb{R}$?  If not, how can I prove that $\int\frac{\vert f-h\vert}{1+\vert f-h\vert}\leq\int\frac{\vert f-g\vert}{1+\vert f-g\vert}+\int\frac{\vert g-h\vert}{1+\vert g-h\vert}$? I tried C-S: $$\sum_{cyc}\frac{|x|}{1+|x|}=\sum_{cyc}\frac{x^2}{|x|+x^2}\geq\frac{(x+y)^2}{x^2+y^2+|x|+|y|}.$$ Thus, it's enough to prove that $$\frac{(x+y)^2}{x^2+y^2+|x|+|y|}\geq\frac{(x+y)^2}{|x+y|+(x+y)^2}$$ or $$|x+y|+(x+y)^2\geq x^2+y^2+|x|+|y|,$$ which is wrong for $(x,y)=(1,-1).$
96526
I have to show the inequality of $$\left(\sum_{i=1}^n x_i\right)*\left(\sum_{i=1}^n \frac{1}{x_i}\right) \geq n^2.$$For $x_1, ... x_n \in \mathbb{R_{&gt;0}}$ and $ n \geq 1$. I wanted to show this inequality by induction. The basis is clear, but I am not sure how to do the inductive step. Can someone help me with that?
68929
Let $x,y,z&gt;0$: Prove that:   $\frac{1}{(x+2y)^2}+\frac{1}{(y+2z)^2}+\frac{1}{(z+2x)^2} \geq\frac{1}{xy+yz+zx}$  I tried to apply Cauchy - Schwarz's inequality but I couldn't prove this inequality!
8478
let $x,y,z&gt;0$ and such $xy+yz+zx=3$,show that $$\dfrac{x}{x^3+y^2+1}+\dfrac{y}{y^3+z^2+1}+\dfrac{z}{z^3+x^2+1}\le 1$$  To prove this inequality,I want use following Cauchy-Schwarz inequality $$(x^3+y^2+1)(\frac{1}{x}+1+z^2)\ge (x+y+z)^2$$ $$\dfrac{x}{x^3+y^2+1}\le \dfrac{1+x+xz^2}{(x+y+z)^2}$$ we have $$\sum\dfrac{x}{x^3+y^2+1}\le\dfrac{3+x+y+z+xz^2+yx^2+zy^2}{(x+y+z)^2}$$ it suffices to prove that  $$x+y+z+xz^2+yx^2+zy^2\le x^2+y^2+z^2+3$$
76504
I'm struggling with proving the following inequality: $$\sum_{i=0}^{t-2}{\frac{1}{t+3i}} \leq \frac{1}{2}$$  for all $t \geq 2$. I think it is monotonic non-increasing in $t$, which would suffice.  Thanks in advance!
109522
If $x_1,x_2,\ldots,x_n&gt;0$ and $x_1+x_2+\ldots+x_n=1$, prove that $$\frac{x_1}{1+x_2+x_3+\ldots+x_n} + \frac{x_2}{1+x_1+x_3+\ldots+x_n} +\ldots + \frac{x_n}{1+x_1+x_2+\ldots +x_{n-1}} \ge \frac {n}{2n-1}$$  This can easily be simplified: $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} \ge \frac {n}{2n-1}$$  I could try using the Cauchy-Schwarz inequality: $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} = \frac{x_1^2}{2x_1-x_1^2} + \frac{x_2^2}{2x_2-x_2^2} +\ldots + \frac{x_n^2}{2x_n-x_n^2} \ge \frac{(x_1+x_2+\ldots+x_n)^2}{2(x_1+x_2+\ldots+x_n)-(x_1^2+x_2^2+\ldots+x_n^2)}=\frac{1}{2-(x_1^2+x_2^2+\ldots+x_n^2)}=\frac{n}{2n-(x_1^2+x_2^2+\ldots+x_n^2)n}$$  It's left to prove that $$(x_1^2+x_2^2+\ldots+x_n^2)n \ge 1$$ or $$(x_1^2+x_2^2+\ldots+x_n^2)n=1$$I can continue using the Cauchy-Schwarz inequality again: $$(x_1^2+x_2^2+\ldots+x_n^2)n=(x_1^2+x_2^2+\ldots+x_n^2)(1+1+\ldots+1)\ge(x_1+x_2+\ldots+x_n)^2=1$$ So I've proved it myself. (I've shown this proof after editing, I didn't post the question with the solution in the details. I found the proof a while after putting the question here).
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?
104491
Can anyone help me with the above inequality? I tried looking at the series expansion and I guess the answer indeed lies there, but I fail to see it.  Thanks
46228
This is used to go from Bennett's inequality to Bernstein's inequality. Yet I don't understand how to prove the inequality.  Assume that $u &gt; 0$, define $$ h(u) = (1+u) \log (1+u) - u $$ show that $$ h(u) \ge \frac{u^2}{2(1+u/3)} $$    My research :  Decomposing the function $h$ as a power series show that it is equivalent to $$ \sum_{n=1}^\infty (-1)^n \frac{u^n}{n+1} \frac{n-1}{n (n+2)} \ge 0 $$ Sadly, I see no reason for this series to be positive.    SOLUTION  The twice differenciation technique given below works. However, I don't agree with the calculus, only on small things that don't change the result. Define $$ g(u) = (1+u)\log (1+u) - u - \frac{3u^2}{2(3+u)} $$ $$ g'(u) = \log (1+u) + 1 - 1 - \frac{3}{2} \frac{2u (3+u) - u^2}{(3+u)^2}\\ $$ Simplifying, $$ g'(u) = \log (1+u) - \frac{3}{2} \frac{u(u+6 )}{(3+u)^2}   $$ $$ g''(u) = \frac{1}{1+u} - \frac{3}{2} \frac{(2u +6)(3+u)^2 -  2 (u+3)(u^2 + 6u) }{(3+u)^4}   $$ Simplifying,  $$ g''(u) = \frac{1}{1+u} - \frac{3}{2} \frac{2u^2 +6u + 6u + 18 -  2u^2 - 12u }{(3+u)^3}   =  \frac{1}{1+u} - \frac{ 27  }{(3+u)^3}  $$  Simplifying again,  $$ g''(u)= \frac{ u^3 + 3\times 3u^2 + 3 \times 9u + 27 - 27(1+u)}{(1+u)(3+u)^3}  = \frac{ u^2(u+9)}{(1+u)(3+u)^3} &gt; 0 $$ And the result is given by the reasoning of Clement.
113910
EDIT:  My question was poorly worded.   I wasn't asking about showing $\ln(1+x) &gt; \frac{2x}{2+x}$ for $x&gt;0$.   What I wanted to know is why the lower bound provided by $ \frac{2x}{2+x}$ was so tight for small positive values of $x$. This is addressed in robjohn's answer.
71084
I have an assignment, that I cannot seem to find a starting point for. I am not looking for the specific answer, but only where to begin. Any pointers are appreciated.  I have to prove the inequality $\vert \tanh(x)-x+\frac{x^3}{3}\vert\leq\frac{1}{8}$ for $x\in[-1/4,1/4]$ where $x-\frac{x^3}{3}$ is the 3rd order Taylor expansion.  I've tried with the triangle inequalities, Lagrange remainders, and a lot of other stuff, but I am no where near a correct argumentation. Does anyone have any suggestions?
114114
For positive real numbers $a,b,c$ with $abc = 8$ prove that   $$ \frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} + \frac{b^2}{\sqrt{(1+b^3)(1+c^3)}} + \frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \frac{4}{3}. $$   Can we prove this by Cauchy-Schwarz or Jensen's inequality? If not how?
5280
I have seen the following inequality  $$ \frac{x - y}{\log x - \log y} &gt; \sqrt{xy} \ , \quad \forall x&gt;y $$ be stated as a near "obvious" fact in another , on the site.  The inequality is very cute, but so far I have not been able to prove it.  It reminds me of the Lipschitz inequality, but has some minor differences. Also Jensens inequality comes to mind. Is there something obvious I am missing, or is this ineqality not as easy to prove as it looks?
47306
How do I prove the following inequality :  $$\Bigg(\frac{2}{\alpha^2} \, \big( e^{\alpha x} - e^{\alpha y} \big) \, + \, e^{\alpha y} (y^2 - x^2) \; \Bigg) &gt; 0 $$  given, $x, y &gt; 0$ ?  Can anyone provide me with hints about this problem ?  Thanks in advance.  Here, $\alpha$ is a positive constant.
16048
mathematicians! I want to ask to all wise people about a problem I met at the quiz to obtain some ideas. The problem is following. It may not be accurate since the problem is dependent on my memory   Let $f : [0, \infty) \rightarrow \mathbb{R}$ and $f$ is differentiable. $|f'(x) |\; \leq\  |f(x)|$ for all $x \in [0, \infty)$. $f(0) = 0$. Then prove that $f(x) = 0$ for all $x \in [0,\infty)$.  At the quiz, I applied the Mean Value Theorem and the Cauchy-Schwarz inequality.   For any $x$, $|f(x)| = |\{f(x) - f(0)\}(x-0)| = |f'(c_{1})(x)|$ for some $c_{1} \in (0,x)$, and then  $$|f'(c_{1})(x)| \leq |f'(c_{1})||x| \leq |f(c_{1})||x|$$ by Cauchy-Schwarz and given assumption. By doing so consecutively, we can obtain the inequality that  $$ |f(x)| \leq |f(c_{n})||x||c_{1}|\ldots |c_{n-1}| $$ for any $x \in [0,\infty)$. As $\{c_{i}\}, \; i\in \mathbb{N}$ is decreasing sequence and $f$ is continuous, $|f(c_{n})|$ can be arbitrary close to $0$ when $c_{n} \rightarrow 0$. Thus the inequality above can be bounded by $\epsilon$.  I think my answer is right, but a bit uncertain. Could you give me a certainty? Thank you very much.
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?