Hello I am a student and had the following problems in inequality: $-2\le 1 $ Squaring both sides the sign changes  $4\gt 1$ But for  $-1\lt 1$ Squaring the terms we get  $1\le 1$ I am stuck here please help me to understand this concept
121268
I want to know what will happen to the inequality in these cases  $a&gt;x&gt;b$ If $a$ and $b$ negative  When i square the both sides of it  Is it going to be  $b^2&gt;x^2&gt;a^2$  Ok but what would happen if  $a$ is a positive and $b$ is a negative number when i square ? EXAMPLE $2&gt;x&gt;-1$  another case  $2&gt;x&gt;-4$   i forgot  mentioning this case ?  If it is something like  $2&gt;x&gt;-2$
91390
how do you prove this inequality? $|\sqrt[n]{x}-\sqrt[n]{y}| \le \sqrt[n]{|x-y|}$ At first glance I thought the triangle inequality would be useful but it's not form what I see. If I raise everything to the power of n I get the following: Possibility 1 (n uneven) : $|x + ... -y| \le |x-y|$ I guess I would have to prove that ... $\le$ 0 but I don't know how to do that. Possibility 2 (n even) : $|x + ... +y| \le |x-y|$ Same as above but now I have to prove that ... $\le$ -y. Am I even on the right track here? Any help would be appreciated. Thanks in advance.
8358
In the pdf which you can download  I found the following inequality which I can't solve it.     Exercise 2.1.11 Let $a,b,c  \gt 0$. Prove that    $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}.$$   Thanks :)
43524
How would I go about proving that  $$|a|+|b| \leq |a+b|+|a-b|$$  Using the triangle inequality?  I tried squaring both sides, yielding:  $$|a|^2 +|b|^2 +2|a||b| \leq 2|a|^2+2|b|^2+2||a|^2-|b|^2|$$  Is it correct to move the terms to the other side? I tried $$2|a||b| \leq |a|^2+|b|^2+2||a|^2-|b|^2|$$ Then it is obvious that  $$0\leq |a|^2+|b|^2-2|a||b| +2||a|^2-|b|^2|$$ $$0\leq (|a|-|b|)^2 +2||a|^2-|b|^2|$$ $$0\leq ||a|-|b||^2 +2||a|^2-|b|^2|$$  But I have neither used the triangle inequality, nor it looks mathematically rigorous for my real analysis class. Any tips and tricks for solving these problems?
4186
Okay, this is my another try on the question  which I unfortunately mis-stated and actually asked for a problem different than the one I have to solve. I wish to show that given the constraints $0 &lt; a_i, b_j &lt; 1$ and $\sum_{i=1}^{n} a_i = \sum_{i=1}^{n} b_i = 1$ we have the following inequality: $$\sum a_i \ln(b_i) \leq \sum a_i \ln(a_i)$$ The problem arose when I was trying to compute some topological pressures and $a_i, b_j$ are actually measures of some sets, but this inequality (if, hopefully, true) is purely algebraic. Can anyone provide some suggestions?
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
91951
Any hints on how I can prove the inequality:     $$|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|},\quad x,y \geq 0$$   Thank you.
95127
I just started doing AM-GM inequalities for the first time about two hours ago. In those two hours, I have completed exactly two problems. I am stuck on this third one! Here is the problem:  If $a, b, c \gt 0$ prove that $$ a^3 +b^3 +c^3 \ge a^2b +b^2c+c^2a.$$  I am going crazy over this! A hint or proof would be much appreciated. Also any general advice for proving AM-GM inequalities would bring me happiness to my heart. Thank you!
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I've been dealing with this problem for almost 2 hours now, with hardly any progress. I'm to prove the following inequality using induction   $$\sqrt[n+1]{n+1}&lt;\sqrt[n]{n}$$ where $n≥3$, $n∈\mathbb{N}$. How can this be done?
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I can show that for $x &gt; 0$ and $r_{i} &gt; 0$ we have $$ \left(\, x + r_{1}\,\right)\ldots\left(\, x + r_{n}\,\right)\ \geq\ \left[\, x + \left(\, r_{1}\ldots r_{n}\,\right)^{1/n}\,\right]^{n}.$$ However, I can't do this using straight up induction, strong or weak. Can someone do this?   You can show the inequality holds for $n$ if it holds for $n+1$ using $r_{n+1} = (r_1 \cdots r_n)^{1/n}$, so going down isn't a problem, but going up has eluded me. Additionally, you can show that if it holds for $n=a$ and $n=b$ then it holds for $n=ab$. Using powers of two and the above is sort of a way to prove the above by induction but it certainly isn't "normal".  You can also prove the above using Lagrange multipliers (which isn't surprising). The last proof I know is where you compare coefficients of $x^k$ and use the AM-GM inequality.
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Prove that $$1&lt;\dfrac{1}{1001}+\dfrac{1}{1002}+\dfrac{1}{1003}+\dots+\dfrac{1}{3001}&lt;\dfrac43 \, .$$  My work: $$\begin{eqnarray*} S&amp;=&amp;\bigg(\dfrac{1}{1001}+\dfrac{1}{3001}\bigg)+\bigg(\dfrac{1}{1002}+\dfrac{1}{3000}\bigg)+\ldots+\dfrac{1}{2001}\\ S&amp;=&amp;\dfrac{1}{4002}\bigg\{\bigg(\dfrac{1001+3001}{1001}+\dfrac{1001+3001}{3001}\bigg)+\ldots\bigg\}+\dfrac{1}{2001}\\ S&amp;\ge&amp; \dfrac{1}{4002}4\cdot1000+\dfrac{1}{2001}=1 \end{eqnarray*}$$ I could derive the left hand inequality with a $\ge$ sign though but could not do anything about the right hand inequality. Please help. EDIT: I do not need to use the equality sign, I can rather use only strict inequality because there exist terms which are not equal so, the AM-HM inequality is actually a strict inequality here. $$S&gt; \dfrac{1}{4002}4\cdot1000+\dfrac{1}{2001}=1$$ Though I got the solution but I am looking for some non-calculus solutions too.
31053
For any $\epsilon&gt;0$ and $a,b\in\mathbb{R}$ show that $$|a-b|&lt;\epsilon\Rightarrow  |a|&lt;|b|+\epsilon$$   In some notes I found the following property $$|a|-|b|\leq ||a|-|b||\leq|a-b|$$ So $$|a-b|&lt;\epsilon\Rightarrow |a|-|b|&lt;\epsilon\Rightarrow |a|&lt;|b|+\epsilon$$  I have two doubts  1) Is it right?  2)In triangular inequality I have that $$|a+b|\leq |a|+|b|$$ $$|a+(-b)|\leq |a|+|-b|\leq|a|+|b|$$  Is wrong do that?
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I am having issues solving these problems: I tried using $(|a|+|b|)^2 \geq 0$ and $(|a|-|b|)^2 \geq 0$ but I am having problems constructing the proof. I need to gain some intuition on how to proceed.     Show that for $a,b\in (-\infty,+\infty)$,   the following inequalities apply:   a) $2|ab| \leq a^2 + b^2$  b) $|a|+|b| \leq \sqrt {2}(a^2+b^2)^{1/2}$
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I am trying to show that, for all $x&gt;0:$  $$\frac{1}{\sqrt{x}}\geq \frac{\ln x}{x-1}$$  This inequality is closer than I expected. I have tried exponentiating, power series, and have achieved nothing. I would really appreciate some help. Below is a graph of the two functions for small $x:$
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I am trying to prove that for $x \in \mathbb{R}^n$ $$\frac{1}{\sqrt{n}}||x||_1 \le ||x||_2 \le ||x||_1$$  So far I have successfully proven that $||x||_2 \le ||x||_1$, which was fairly easy.  I took the square of both norms and used that to show that $||x||_2^2 \le ||x||_1^2$.  The needed result then follows easily.  Where I am getting stuck is in proving that $\frac{1}{\sqrt{n}}||x||_1 \le ||x||_2$.  I took the square of both norms again, which results in $$(\frac{1}{\sqrt{n}}||x||_1)^2 = \frac{1}{n} \sum_{i=1}^n|x_i|^2+\frac{2}{n}\sum_{i \neq j}|x_i||x_j|$$ and $$(||x||_2)^2 = \sum_{i=1}^n|x_i|^2$$  My next step was to prove that $\frac{2}{n}\sum_{i \neq j}|x_i||x_j|$ must be less than the difference between $\frac{1}{n} \sum_{i=1}^n|x_i|$ and $\sum_{i=1}^n|x_i|$, meaning that $$\frac{2}{n}\sum_{i \neq j}|x_i||x_j| \le \frac{n-1}{n}\sum_{i=1}^n|x_i|$$ Which is where I am now stuck.  Is there some identity that I am missing here that will make this easier?
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Here there is my proof (quite short and easy) of a rather straightforward result. The text of this question comes from a previous question of mine, where I ended up working on a wrong inequality. Here things are fixed. Still, I would like to know:  if it is sound, because absolute value always creates me some problem, and if there is a shorter (neater) way to get the result (maybe without using contradiction).    Proposition: $|d(x,y)-d(y,z)| \leq d(x,z)$. Proof: We proceed by cases.  Case 1: $d(x,y)-d(y,z) \geq 0$. Assume by contradiction that $d(x,y)-d(y,z) &gt; d(x,z)$. Hence, $d(x,y)&gt; d(y,z) + d(x,z)$, contradicting the triangle inequality.  Case 2: $d(x,y)-d(y,z) &lt; 0$. Assume by contradiction that $ - d(x,y) + d(y,z) &gt; d(x,z)$. Hence, $d(x,y) &lt; d(y,z) - d(x,z)$. By applying the triangle inequality, we have that $d(y,x) \geq d(y,z) + d(x,z)$, which is by symmetry equivalent to  $d(x,y) \geq d(y,z) + d(x,z)$, obtaining the desired contradiction. $\square$     As always, any feedback is welcome. Thank you for your time.
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Simplify: $x^2 &gt; 1$.  My solution: Taking square root on both sides:  $±x &gt; ±1$  So my results are:   $x &gt; 1$ $x &gt; -1$ $-x &gt; 1$ $\implies$ $(-1 &gt; x)$ $-x &gt; -1$ $\implies$ $(1 &gt; x) $   But I strongly feel this is wrong. What is wrong here?  A step-by-step explanation will help me.