I am trying to prove $|e^z - e^\omega| \leq |z - \omega|$ for $z,\omega\in\{z\in\mathbb{C}\;:\;\Re(z)&lt;0\}$ and I get stuck in an inequality of the stated in the title : $|e^x + e^y| \leq |x - y|$ for $x,y&lt;0$.  This is what I have tried: \begin{equation}\left|e^z - e^\omega\right| = \left|e^{i\Im(z)}\right|\left|e^{\Re(z)} - e^{\Re(\omega)}e^{i(\Im(\omega) - \Im(z))}\right| \leq \left|e^{\Re(z)} + e^{\Re(\omega)}\right| \end{equation}  Is it true that $|e^x + e^y| \leq |x - y|$ for $x,y&lt;0$? If yes, any help will be appreciated.  Thank you!
32648
Came across this problem on an old qualifying exam: Let $a$ and $b$ be complex numbers whose real parts are negative or 0. Prove the inequality $|e^a-e^b| \leq |a-b|$.  If $f(z)=e^z$ and $z=x+iy$, then $|f&#39;(z)|=e^x\leq 1$ given that $x \leq 0$. I played around with the limit definition of the derivative, but wasn't able to get anywhere. Not sure what else to try; a hint would be very helpful!
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?
97775
Find all values of $\omega \in \mathbb{C}$ such that $\frac{\omega - \frac1\omega}{2i} = 2.$  So what I have is:  Let $\omega$ be represented as $x + yi$ (by definition of complex numbers)  Then $\frac1\omega = \frac{x-yi}{x^2 + y^2}$ (by properties of complex numbers)  So, $\frac{x+yi - \frac{x-yi}{x^2 + y^2}}{2i} = 2$  Eventually, I get, via expansion and simplification:  $\frac{x(x^2 + y^2 - 1) + yi(x^2 + y^2 + 1)}{2i(x^2 + y^2)}$  Any hints on how I can continue? I feel like I'm close but missing something.
63657
Prove that for all $x,y&gt;0$ the following inequality $x^xy^y \geq (\frac{x+y}{2})^{x+y}$ is true.  It smells like Jensen inequality, but all I can get is that $\frac{x+y}{2}ln(x) + \frac{x+y}{2} ln(y) \geq xln(\frac{x+y}{2})+yln(\frac{x+y}{2})$
22071
I did substitute $a=x+y, b=x+z, c=y+z$ and I arrived at $\sqrt{2x} + \sqrt{2y} + \sqrt{2z} \le \sqrt{x+y} + \sqrt{x+z} + \sqrt{y+z}$. However, after this, I tried various methods like AM-GM and Cauchy-Schwarz inequality for hours and I still can't prove it. Can someone help please? Thanks.
2180
Show that $|x-y|\leq |x|+|y|$.   I know this probably applies the triangle inequality in some way, but I can't figure it out.
64023
let $x,y,z,w&gt;0$,show that $$(x+y)^3+(y+z)^3+(z+w)^3+(w+x)^3\ge 8(x^2y+y^2z+z^2w+w^2x)$$  it seem use AM-GM inequality to solve it,But I can't it,Thanks
30772
Suppose $z,z_1,z_2,z_3 \in \mathbb{C}$ such that $z^2 = z_1^2+z_2^2+z_3^2$.  Prove that $$|\Re(z)| \le|\Re(z_1)|+|\Re(z_2)|+|\Re(z_3)|$$ where $\Re(z)$ is the real part of $z$.  I think we can use the triangle inequality but I can't complete it.
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?
103710
Let $f,g : \Omega \subseteq \mathbb R^n \rightarrow [0,+\infty]$ be measurable functions with $f(x) = g(x)$ a.e. . Then I have to show that $\int_\Omega  f = \int_\Omega g$.  I may not assume that $\int_\Omega (f+g) = \int_\Omega f + \int_\Omega g$.   This task is from Tao Proposition 19.2.6.
91198
Let $x&gt;0$, $y&gt;0$ and $z&gt;0$. Show that   $$x^3y+y^3z+z^3x\ge xyz(x+y+z).$$   I known we can't WLOG: $x\ge y\ge z$, if this, I can use rearrangement inequality, But other I can't it. Thanks?
38566
Prove that $\forall x,y&gt;0, x^x+y^y \geq x^y + y^x$   A friend of mine told me none of the teachers in my school have succeeded in proving this seemingly simple inequality (it was asked at an oral exam last year). I tried it myself, but I made no significant progress.
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  $$x+y+z+xz^2+yx^2+zy^2\le x^2+y^2+z^2+3$$
85884
Show that for $x,y,z &gt; 0$ the inequality is true: $\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+x+y+z \geq \frac{(x+y)^2}{y+z}+\frac{(y+z)^2}{z+x}+\frac{(z+x)^2}{x+y}$  I have tried Holder, but i had no luck. Please give me a hint of how to start.
63727
I tried to search for a few minutes but I didn't find this question so I hope it's not a duplicate.  So I want to show that $(e^x)' = e^x$. To do that, I must proof that the limit: $$\lim_{h\to0}\frac{f(x+h) - f(x)}{h} = \lim_{h\to0}\frac{e^{x+h} - e^x}{h}$$ exists and equals to $e^x$.    So I have: $$\frac{e^{x+h} - e^x}{h} = \frac{e^x \cdot e^h - e^x}{h} = e^x \bigg(\frac{e^h - 1}{h}\bigg) \\ e^h - 1 = z \implies e^h = 1+z &gt;0 \implies h = \ln(1+z)$$ Because of the continuity of the $\ln$ and $e^x$ functions, we have: $$h\to 0 \iff e^h \to 1 \implies z\to0 \\ z &gt; 0 \implies \frac{1}{z} \to +\infty \\ z &lt; 0 \implies \frac{1}{z} \to -\infty \\ \frac{e^h - 1}{h} = \frac{z}{\ln(1+z)} = \frac{1}{\frac{1}{z}\ln(1+z)} = \bigg[ y = \frac{1}{z} \bigg] =  \frac{1}{y\ln(1+\frac{1}{y})} = \frac{1}{\ln(1+\frac{1}{y})^y} \\ h \to 0 \implies |y| \to +\infty \implies \bigg(1+\frac{1}{y}\bigg)^y \to e \implies \\ \lim_{h\to0}\frac{e^h - 1}{h} = \frac{1}{\ln e} = 1 \implies (e^x)' = e^x \cdot 1 = e^x$$    But how do I prove that $\lim_{\pm\infty}(1 + \frac{1}{x})^x = e$ ?    We defined $e$ like this: $$\lim_{n\to\infty} \bigg(1 + \frac{1}{n}\bigg)^n = e \space \,, \space n\in\mathbb{N}$$   I thought of the sandwich theorem:  $\forall x \geq 1 \,, x\in\mathbb{R} \,, \lfloor x \rfloor \leq x &lt; \lfloor x \rfloor + 1$ and $\lfloor x \rfloor , \lfloor x \rfloor + 1 \in \mathbb{N} $ and if I denote $\mathbb{N} \ni n = \lfloor x \rfloor$ I have: $$ \bigg(1 + \frac{1}{n+1}\bigg)^n \leq  \bigg(1 + \frac{1}{x}\bigg)^x \leq \bigg(1 + \frac{1}{n}\bigg)^{n+1} \\   \lim_{n\to+\infty}\bigg(1 + \frac{1}{n+1}\bigg)^n = \frac{\lim_{n\to+\infty}\bigg(1 + \frac{1}{n+1}\bigg)^{n+1}}{\lim_{n\to+\infty}\bigg(1 + \frac{1}{n+1}\bigg)} = \frac{e}{1+0} = e \\  \lim_{n\to+\infty}\bigg(1 + \frac{1}{n}\bigg)^{n+1} =  \lim_{n\to+\infty}\bigg(1 + \frac{1}{n}\bigg)^n \cdot \lim_{n\to+\infty}\bigg(1 + \frac{1}{n}\bigg) = e \cdot (1 + 0) = e$$  and now I could apply the theorem but then again, how do I show that this: $\bigg(1 + \frac{1}{n+1}\bigg)^n \leq  \bigg(1 + \frac{1}{x}\bigg)^x \leq \bigg(1 + \frac{1}{n}\bigg)^{n+1}$ is true? I mean, I can say that (left side): $$ n+1 &gt; x \implies \frac{1}{n+1} &lt; \frac{1}{x} \implies 1 + \frac{1}{n+1} &lt; 1 + \frac{1}{x} \\ \forall y\in\mathbb{R_+} \space n \leq x \implies y^n \leq y^x \\ \implies \bigg(1 + \frac{1}{n+1}\bigg)^n \leq  \bigg(1 + \frac{1}{x}\bigg)^x$$ and by analogy the right side but is this a proof in terms of limits?  and that's just $+\infty$ for $x \geq 1$, what do I do for $-\infty$ ?   For $x\in\langle 0, 1\rangle$, a substitution $y = \frac{1}{x}$ gives me: $$\lim_{y\to 0}\bigg(1 + y\bigg)^\frac{1}{y}$$  I'll stop here to see if I'm on the right track, any hints, suggestions, edits, comments and answers are welcome!
57864
Can somebody please check if my proof is okay?  Prove that $||z_1|-|z_2|| \leq |z_1-z_2|$.  Attempt: This proof seems similar to the triangle inequality proof.  $$\begin{split} |z_1-z_2|^2 &amp;= (z_1-z_2) \overline{(z_1-z_2)} \\ &amp;= (z_1-z_2)(\bar z_1 - \bar z_2) \\ &amp;= z_1\bar z_1 - z_1 \bar z_2 - \bar z_1 z_2 + z_2 \bar z_2\\ &amp;=|z_1|^2 - 2Re(z_1 \bar z_2) + |z_2|^2\\ &amp; \geq |z_1|^2 -2|z_1 \bar z_2| + |z_2|^2 \\ &amp;= |z_1|^2 - 2|z_1||z_2| + |z_2|^2 \\ &amp;= (|z_1| - |z_2|)^2. \end{split}$$  Taking the square root of both sides gives $|z_1|-|z_2| \leq |z_1-z_2|$  Can I just take the norm of both sides so that $||z_1|-|z_2|| \leq ||z_1-z_2|| = |z_1-z_2|$?  Thank you.
1837
Consider following inequality:     $|z_1 + z_2|\ge|z_1| - |z_2|  \  \ \ \ \ \  z_1,z_2 \in \mathbb{C}$   Is that true always? I tried many examples and it was correct but I don't know how to prove that if it's right.