Articles of inequality

Inequality for $\sum_{n=1}^\infty \frac{x^n}{n^n}$.

I need to show that for $x<0$, $$\sum_{n=1}^\infty \frac{x^n}{n^n}<0$$ but I am completely stuck. I noted that the series is alternating, the first term is negative, but the term is only eventually decreasing. Any hint?

Proving an inequality about a sequnce with Cauchy-Schwarz

show that $$\sum\limits_{i=1}^n \frac{x_i}{i^2} \geq \frac{1}{1} + \frac{1}{2} + \dots +\frac{1}{n}$$ where $x_1,x_2,\dots,x_n$ are natural numbers and all of them are different numbers(no such a $x_i=x_j$) the teacher said you can prove it by making it a Cauchy form inequality. thing i have tried to make Cauchy inequality and show it’s same as question inequality: […]

Prove that for every positive integer $n$, $1/1^2+1/2^2+1/3^2+\cdots+1/n^2\le2-1/n$

Base case: n=1. $1/1\le 2-1/1$. So the base case holds. Let $n=k\ge1$ and assume $$1/1^2+1/2^2+1/3^2+\cdots+1/k^2\le 2-1/k$$ We want to prove this for $k+1$, i.e. $$(1/1^2+1/2^2+1/3^2+\cdots+1/k^2)+1/(k+1)^2\le 2-\frac{1}{k+1}$$ This is where I get stuck. Any help appreciated.

Prove this inequality $\frac{1}{1+a}+\frac{2}{1+a+b}<\sqrt{\frac{1}{a}+\frac{1}{b}}$

Let $a,b>0$ show that $$\dfrac{1}{1+a}+\dfrac{2}{1+a+b}<\sqrt{\dfrac{1}{a}+\dfrac{1}{b}}$$ It suffices to show that $$\dfrac{(3a+b+3)^2}{((1+a)(1+a+b))^2}<\dfrac{a+b}{ab}$$ or $$(a+b)[(1+a)(1+a+b)]^2>ab(3a+b+3)^2$$ this idea can’t solve it to me,are we aware of an elementary way of proving that? Thanks in advance.

Is the derivative of a function bigger or equal to $e^x$ will always be bigger or equal to the function ?!

It seems to be the case, but i don’t have a proof. Given the function $f$ such that $f(x) \geq e^x$, is it true that $f'(x) \geq f(x)$?! I was experimenting with wolfram and it appears that $\frac{f'(x)}{f(x)} \geq 1$ whenever $f$ is bigger or equal to $\exp(x)$. Note : as suggested in the comments, […]

Prove that $\forall x>0, \frac {x-1}{\ln(x)} \geq \sqrt{x} $.

This inequality arose in this question Prove that : $|f(b)-f(a)|\geqslant (b-a) \sqrt{f'(a) f'(b)}$ with $(a,b) \in \mathbb{R}^{2}$ : $$\forall x>0, \frac {x-1}{\ln(x)} \geq \sqrt{x} $$ Can anybody find a way to prove it without calculus? Like using only inequalities like AM-GM, Jensen or Cauchy – Schwarz? CLARIFICATION I want a proof that does not involve […]

Proving AM-GM via $n \cdot (a_1^n + a_2^n + \dots + a_n^n) \ge (a_1^{n-1} + a_2^{n-1} + \dots + a_n^{n-1}) \cdot (a_1 + a_2 + \dots + a_n)$

I want to prove the arithmetic–geometric mean inequality. To prove that, I need the following inequality: Suppose that $n$ is an integer which is greater than or equal to $1$ and $a_1, a_2, \dots, a_n \in \Bbb{R}$. Then, $$n \cdot (a_1^n + a_2^n + \dots + a_n^n) \ge (a_1^{n-1} + a_2^{n-1} + \dots + a_n^{n-1}) […]

with inequality $\frac{1}{3a+5b+7c}+\frac{1}{3b+5c+7a}+\frac{1}{3c+5a+7b}\le\frac{\sqrt{3}}{4}$

let $a,b,c>0$, such $ab+bc+ac=1$,show that $$\dfrac{1}{3a+5b+7c}+\dfrac{1}{3b+5c+7a}+\dfrac{1}{3c+5a+7b}\le\dfrac{\sqrt{3}}{4}$$ by Macavity C-S:with inequality $\frac{y}{xy+2y+1}+\frac{z}{yz+2z+1}+\frac{x}{zx+2x+1}\le\frac{3}{4}$ $$\dfrac{1}{3a+5b+7a}\le\dfrac{1}{2}\left(\dfrac{1}{6a+9b}+\dfrac{1}{b+7c}\right)$$ It suffices to show $$\sum_{cyc}\left(\dfrac{1}{6a+9b}+\dfrac{1}{b+7c}\right)\le\dfrac{\sqrt{3}}{2}$$

Will an increasing function applied inside definite integrals preserve inequality?

If $$\int_a^b f(x) \,dx > \int_a^b g(x) \,dx$$ and there is a function $h(x)$ that is strictly increasing with $x$, does that imply that $$\int_a^b h(f(x)) \,dx > \int_a^b h(g(x)) \,dx$$ ?

Proof of inequality without calculus

So we are given the equation $3x+4y+xy=2012$ where $x$ and $y$ are positive integers. Prove that $x+y\geq83$. Using calculus optimisation methods, this can be proved. However, it requires a lot of difficult calculations, and students are not expected to know calculus to answer this. How else could you prove this? You could always go through […]