Tag Info

Hot answers tagged

17

Some general properties of the absolute value First let me recall that $$ |x| :=\begin{cases}x & \text{if }x \geq 0,\\ -x & \text{if }x <0.\end{cases}$$ We show that $|x|=|-x|$ for every $x\in \Bbb R$: If $x\geq 0$ then $|x| = x$ and $|-x|= -(-x)=x$. If $x< 0$ then $|x| = -x$ and $|-x|= -x$. $$ \rightarrow \qquad \boxed{\quad |x|=|-x| ...


9

How about the substitution $x = \tan \theta$, then the inequality reduces to proving: $$\sin \tan \theta \ge \sin \theta$$ for, $\theta \in \left(0,\tan^{-1} (\pi/2\right))$. Since, $\sin \theta$ is monotone increasing in the interval $\left(0,\pi/2\right)$, we are done using: $$\tan \theta \ge \theta$$


9

Apply AM-GM inequality to each factor: $$p+2\ge2\sqrt{2p}$$ $$q+2\ge2\sqrt{2q}$$ $$p+q\ge2\sqrt{pq}$$ Multiply the three inequalities and you are done.


9

Your differentiating is not valid. Comparison of gradients is not enough for comparison of functions. I'll provide two alternatives to prove it below: Taylor Expansion: Using the Taylor expansion of $e^x$, $$e^x=1+x+\frac{x^2}{2!}+ \cdots$$ so that $e^x > 1 + x$, as long $x \neq 0$. Then letting $x = \frac{\pi}{e} - 1$ we have $$e^{\pi/e -1} > ...


6

You cannot differentiate like that. That's very wrong. Hint: consider the function $f(x)=x^{1/x},\space x>0$. Show that $f$ attains its maximum when $x=e$. And you are done.


6

No, your proof is invalid. The idea is good, though: the inequality $$ x^e<e^x $$ is equivalent (for $x>0$) to $e\log x<x$. Consider the function $$ f(x)=x-e\log x $$ defined for $x>0$. Its limits at $0$ and at $\infty$ are both $\infty$. Next you can consider the derivative: $$ f'(x)=1-\frac{e}{x}=\frac{x-e}{x} $$ which shows the function is ...


6

$\arcsin$ is an increasing function, and you can rewrite $$x>\arcsin\left(\frac{x}{\sqrt{x^2+1}}\right).$$ Then, deriving (and using equality at $x=0$) $$1>\frac{\frac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2+1}}}{x^2+1}}{\sqrt{1-\left(\frac{x}{\sqrt{x^2+1}}\right)^2}}=\frac1{x^2+1}.$$ The original inequation is a mixture of trigonometric and algebraic ...


6

Square both sides (assume $x\in(0,\pi/2)$): $$\sin x>\frac{x}{\sqrt{x^2+1}}\iff 1-\cos^2 x>1-\frac{1}{x^2+1}$$ $$\iff \cos ^2 x<\frac{1}{x^2+1}=\frac{\sin^2 x+\cos^2 x}{x^2+1}$$ $$\iff x^2\cos^2 x+\cos^2 x<\sin^2 x + \cos^2 x$$ $$\iff \tan^2 x> x^2\iff \tan x>x,$$ which is true ($(\tan x-x)'>0$ for $x\in(0,\pi/2)$ and ...


6

First of all, we need to have $x+7\ge 0$. Then, note that $\sqrt{x+7}$ is non-negative. So, since we have $$0\le\sqrt{x+7}\lt x,$$ we have $$0\lt x.$$ Hence, we have $$x+7\ge0\ \ \ \text{and}\ \ \ x+7\lt x^2\ \ \ \text{and}\ \ \ 0\lt x.$$


6

The intuition is that if $g(x) \geq h(x) ~\forall x \in \mathbb R$, then $E[g(X)] \geq E[h(X)]$ for any random variable $X$ (for which these expectations exist). This is what one would intuitively expect: since $g(X)$ is always at least as large as $h(X)$, the average value of $g(X)$ must be at least as large as the average value of $h(X)$. Now apply ...


6

Hint :$x+3\sqrt[3]{xy^2}=x+\sqrt[3]{xy^2}+\sqrt[3]{xy^2}+\sqrt[3]{xy^2}\ge 4\sqrt[4]{x\cdot\sqrt[3]{xy^2}\cdot\sqrt[3]{xy^2}\cdot\sqrt[3]{xy^2}}=4\sqrt{xy}$


5

Case 1 is not for $x>0$ but for $5x-3>0\implies x>\frac{3}{5}$ So for case 1 you have $\frac{3}{5}<x<3+\sqrt{2}$ (since $\frac{3}{5}>3-\sqrt{2}$ For case 2 you have $x<\frac{3}{5}$, so $-5<x<\frac{3}{5}$ So the general solution is $-5<x<3+\sqrt{2}$


5

By AM-GM inequality, since $x,y,z>0$: $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \geq 3\sqrt[3]{\frac{x}{y}\frac{y}{z}\frac{z}{x}}=3$$ with equality iff $\frac{x}{y}=\frac{y}{z}=\frac{z}{x}$, i.e. $x=y=z$. Alternatively, rearrangement inequality. $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ is cyclic - two cases: wlog $x\ge y\ge z$. Then ...


5

Another way is to note the inequality is equivalent to $$(x-6)^2>(x^2-5x+9)^2\iff -(x-1)(x-3)(x^2-6x+15)>0$$ The quadratic is always positive, so this is the same as $(x-1)(x-3)<0$ which means $x\in (1,3)$.


4

Use Cauchy-Schwarz inequality we have $$3(c^2+d^2+e^2)\ge (c+d+e)^2\Longrightarrow 48-3a^2-3b^2\ge (8-a-b)^2$$ then let $2a+b=t$, you have $$8a^2+(8-7t)a+2t^2-8t+8\le 0$$ $$\Longrightarrow \Delta_{a}=(8-7t)^2-32(2t^2-8t+8)\ge 0\Longrightarrow \dfrac{8}{5}\le t\le 8$$


4

Here's what the author is using: Suppose $\{x_n\}$ and $\{y_n\}$ are sequences that converge to $x$ and $y$ respectively and for all $n$, $x_n \leq y_n$. Then $x \leq y$. The way we get around the example in your question is to avoid the strict inequality. Things do go awry when use $<$. However, with $\leq$, things do not. A proof of the statement, for ...


4

Here is an analytic proof inspired by the one in the answer to question #1343375 (thanks to kubek789 for the link!). We fix a positive integer $n$. We let $\left[ n\right] $ denote the set $\left\{ 1,2,\ldots,n\right\} $. We first show an auxiliary result: Theorem 1. Let $A_{1},A_{2},\ldots,A_{n}$ be $n$ finite nonempty sets. Let $x\in\left( ...


4

Use Cauchy-Schwarz inequality we have $$\left(\dfrac{1}{1+a}+\dfrac{2}{1+a+b}\right)^2\le\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(\dfrac{a}{(1+a)^2}+\dfrac{4b}{(1+a+b)^2}\right)$$ so suffices to show that $$\dfrac{a}{(1+a)^2}+\dfrac{4b}{(1+a+b)^2}<1$$ since $$\dfrac{4b}{(1+a+b)^2}<\dfrac{4b}{4(1+a)b}=\dfrac{1}{1+a}$$ it suffices to show that ...


4

If $|a-b|<5$, then $|b|=|b-a+a|\le |b-a|+|a| < 5+|a|$. That's it!


4

suppose $n=k, \\ \dfrac{1}{1+a_{1}}+\dfrac{2}{1+a_{1}+a_{2}}+\cdots+\dfrac{k}{1+a_{1}+a_{2}+\cdots+a_{k}} \\ \le\dfrac{k}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}}$ when $n=k+1 $ LHS=$\dfrac{1}{1+a_{1}}+\dfrac{2}{1+a_{1}+a_{2}}+\cdots+\dfrac{k}{1+a_{1}+a_{2}+\cdots+a_{k}}+\dfrac{k+1}{1+a_{1}+a_{2}+\cdots+a_{k+1}} ...


4

The problem in your proof is that you are assuming that $x+3>0$, which is not true. To solve it, you put everything above the same denominator, keeping in mind that $x \not =3$ and $x \not = -5$. So $\dfrac{(4x+19)(x-3)-(4x-17)(x+5)}{(x+5)(x- 3)}<0$ hence $\dfrac{4(x+7)}{(x-5)(x+3)}<0$ so $(x+5)(x-3)<0$ and $x+7>0$ or $(x+5)(x-3)>0$ ...


3

We want: $$ (m+2)(e^{-x}+1)^2> 2 \tag{1}$$ hence it is enough to compute the infimum of $(e^{-x}+1)^2$ over $\mathbb{R}$. It is trivially $1$, so $(1)$ holds for any $m\geq 0$.


3

Hint: Note that $\sqrt{x+7}$ is a real number only if $x+7\ge 0$ and, in this case,$\sqrt{x+7}$ is a positive number so that we must have $x \ge 0$. From all these conditions you find : $$ \sqrt{x+7}>0 \iff \begin {cases} x\ge 0\\ x+7<x^2 \end{cases} $$


3

Consider $x_i = \alpha_i^{\frac{1}{2}}$ and $y_i = \alpha_i^{-\frac{1}{2}}$ for $i = 1,2,\ldots,n$. By Cauchy-Schwarz inequality, $$\vert \langle x, y \rangle \vert \leq \|x\|_2\|y\|_2$$ which is $$\vert \sum_{i=1}^n x_i y_i \vert^2 \leq (\sum_{i=1}^n x_i^2)(\sum_{i=1}^n y_i^2)$$ and therefore $$n^2 = \vert \sum_{i=1}^n ...


3

$e^{x/e}$ is convex and its tangent at $x=e$ is $y = x$, hence $e^{x/e} \ge x$


3

As others have pointed out, you cannot differentiate and think the inequality remains valid. However, you can integrate, retracing your steps. $$x>e\implies \frac1x<\frac1e\implies \int_e^{\pi}\frac1x\,dx<\int_e^{\pi}\frac1e\,dx\implies \log \pi < \frac{\pi}e\implies \pi^e<e^\pi$$


2

Hint: Show that $|1 - (1/2)^{1/\sqrt{n}}| \leqslant 2/\sqrt{n}$ using the Bernoulli inequality.


2

This is an application of a basic theorem called squeeze theorem: $\def\nn{\mathbb{N}}$ Given any sequences $(a_n)_{n\in\nn}$ and $(b_n)_{n\in\nn}$ and $(c_n)_{n\in\nn}$ such that $a_n \le b_n \le c_n$ for any $n\in\nn$:   If $\lim_{n\to\infty} a_n = \lim_{n\to\infty} c_n = x$ for some $x$:     $\lim_{n\to\infty} b_n = x$. ...


2

Its a pretty standard theorem which was originally called "Sandwich Theorem" but nowadays more popularly known as "Squeeze Theorem". Squeeze Theorem: If $\{a_{n}\}, \{b_{n}\}, \{c_{n}\}$ are sequences such that $a_{n} \leq b_{n} \leq c_{n}$ for all $n$ greater than some specific positive integer $N$ and $\lim_{n \to \infty}a_{n} = \lim_{n \to \infty}c_{n} = ...


2

Let we set $(a,b,c,d,e)=\frac{8}{5}\cdot(1,1,1,1,1)+(x_1,x_2,x_3,x_4,x_5)$. Then we have: $$\left\{\begin{array}{rcl}x_1+x_2+x_3+x_4+x_5&=&0,\\ x_1^2+x_2^2+x_3^2+x_4^2+x_5^2&=&\frac{16}{5}\end{array}\right.$$ and we have to find the stationary points of $2x_1+x_2$. Lagrange multipliers give: $$ (2,1,0,0,0) = ...



Only top voted, non community-wiki answers of a minimum length are eligible