Suppose $f,g,h&gt;0$ for every $x$. How can I show this inequality $$ \frac{|f(x)-g(x)|}{1+|f(x)-g(x)|}\le \frac{|f(x)-h(x)|}{1+|f(x)-h(x)|}+\frac{|h(x)-g(x)|}{1+|h(x)-g(x)|} $$ Adding subtracting $h(x)$ to numerator is the way to go for numerator part but what about denominator?
121436
Show $\rho (x,y)=\dfrac{d(x,y)}{1+d(x,y)}$ is a metric on the metric space $X$, equipped with the Euclidean metric $d$.   I've already shown that the positivity $\rho(x,y)\geq 0$, the symmetry $\rho(x,x)=\rho(y,y)=0$.  I'm having trouble proving the last condition of a metric space though: the triangle inequality must be valid:  $\rho(x,y)\leq \rho(x,p)+\rho (p,y)$, for $p\in X$. So, $\dfrac{d(x,y)}{1+d(x,y)}\leq \dfrac{d(p,x)}{1+d(p,x)}+\dfrac{d(p,y)}{1+d(p,y)}$  Could someone provide a hint as to get started? No complete solutions please. Thanks in advance.
102391
Question: Prove that $\sin x+2x\ge\frac{3x(x+1)}{\pi}$ for all $x\in [0,\frac{\pi}{2}]$  How did $\pi$ come in the expression?  This is how I tried to solve.  $\sin x\le x$ so $\sin 2x\le 2x$  $\sin x+2x\le 3x$.  Applying AM GM inequality: $\sin x+2x\ge 2\sqrt{2x\sin x}$  $3x\ge 2\sqrt{2x\sin x}$  This how far I can go.
44373
I am wondering how to prove both of this by induction: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2^n}\leq 1+\frac{n}{2}$ In general, how should I think of how to prove such an inductive step for these types of inequalities? If there is any other way to prove please feel free. Any help would be appreciated.
102433
This is a pretty straightforward question. I want to show $(x+y)^a &gt; x^a + y^a$ for $x,y&gt;0$ and $a&gt;1$.  One way would be this. WLOG, suppose $x \leq y$. Then:  $(1+\frac{x}{y})^a &gt;1+\frac{x}{y} \geq 1+(\frac{x}{y})^a$.  Noting that $\frac{1}{y^a}t_1&gt;\frac{1}{y^a}t_2 \implies t_1 &gt; t_2$, it follows that $(x+y)^a &gt; x^a + y^a$.  But what other ways are there to show this inequality?
90616
How can I prove this inequality for all natural numbers $n\geq 2$?  ${2n\choose n}&gt;\frac{4^n}{n+1}$  I've tried induction but that was a dead end.
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?
28911
Suppose $f(x)$ and $g(x)$ are continuous function from $[0,1]\rightarrow [0,1]$, and $f$ is monotone increasing, then how to prove the following inequality: $$\int_{0}^{1}f(g(x))dx\le\int_{0}^{1}f(x)dx+\int_{0}^{1}g(x)dx$$
76185
How can I prove that this function is bounded using the given 2 hints? $$f(x) = \frac{x}{x^2 + 1},$$  Hints:   $(x + 1)^2 \geq 0.$ $(x - 1)^2 \geq 0.$   thanks!
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.   $$\frac 1{x+1}&lt;\ln\left(\frac{x+1}x\right)&lt;0.5\left(\frac 1x+\frac 1{x+1}\right)$$
104836
I have to prove the following inequality:     $$\forall n,m \in \mathbb{N},i\in\mathbb{N_0}:n&lt;m \implies \frac1{n^i}\binom{n}i \leq\frac1{m^i}\binom{m}i $$   I used this fact to prove an other inequality, without proving this one first, but now I am stuck. Is induction the way to do this?
116810
How to show  $x_1,x_2, \dots ,x_n \geq 0 $ and $ x_1 + x_2 + \dots + x_n \leq \frac{1}{2} \implies (1-x_1)(1-x_2) \cdots (1-x_n) \geq \frac{1}{2}$
28853
Prove the following inequality:   $$\dfrac{e^x + e^{-x}}2 \le e^{\frac{x^2}{2}}$$   This should be solved using Taylor series.  I tried expanding the left to the 5th degree and the right site to the 3rd degree, but it didn't help.  Any tips?
78664
I'd like to prove that $$\frac{(a-b)^2}{8a}\leqslant\frac{a+b}{2} - \sqrt{ab}\leqslant\frac{(a-b)^2}{8b}$$ when $ 0&lt;b\leqslant a\,$. I have been brushing up on inequalities and I cannot figure this one out. I have searched this site, and found this  , but haven't seen how to apply it. I have tried to find a factor that I can multiply the entire inequality with that will allow me to simplify the middle term. So far no luck there. I have also tried assuming that $b &gt; a$ and looked for a contradiction to either of the terms individually. Does anyone have any hints that can get me somewhere?
103888
I am working on proving the below inequality, but I am stuck.  Let $g$ be a differentiable function such that $g(0)=0$ and $0&lt;g'(x)\leq 1$ for all $x$. For all $x\geq 0$, prove that $$\int_{0}^{x}(g(t))^{3}dt\leq \left (\int_{0}^{x}g(t)dt  \right )^{2}$$
75435
I'm learning calculus, specifically derivatives and applications of MVT, and need help with the following exercice:      Show that, for all $n &gt; 1$ $$\frac{1}{n + 1} &lt; \log(1 + \frac1n) &lt; \frac1n.$$    I tried to follow the below steps in order to prove the RHS inequality:   Proving that $f &lt; g$ on $I$ from $a$ to $b$:  Step $1$. Prove that $f' &lt; g'$ on $\operatorname{Int}(I)$.  Step $2$. Show that $f(a) \leq g(a)$ or that $f(a^+) \leq g(a^+)$  Following the above steps, let $f(x) = \log(1 + \frac1x)$ and $g(x) = \frac1x$, for all $x &gt; 1$. One has  $$f'(x) =  -\frac{1}{x^2 + x} \quad \text{and} \quad g'(x) = -\frac{1}{x^2}.$$  We note that, for every $x &gt; 1$, $f'(x) &gt; g'(x)$. Moreover, $f(1^+) = \log(2) &lt; 1 = g(1^+).$   My problem is that I got the wrong inequality sign in Step $1$.    Looking at the solution in my textbook, the author suggests using the MVT but I don't know how to apply it in this case.
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: