$|Z_1| = | \frac{v(1+\alpha) + \sqrt{v^2(1+\alpha)^2-4\alpha}}{2}|$  Triangle inequality |x+y|=|x|+|y|  Where x= $\frac{v(1+\alpha)}{2}$ and $y= \frac{\sqrt{v^2(1+\alpha)^2-4\alpha}}{2}$  I've been stuck on this problem now for a couple of days and I'm having a difficult time proving this case.   I'm trying to prove that $|Z_1| \leq 1$ while using triangle inequality.   Where in Case I: $−1\leq \alpha \leq 0$ and $0&lt;v&lt;1$  I was wondering if anybody can assistance me on this problem. I want to thank you ahead of time for your cooperation.
122294
$|Z_1| = | \frac{v(1+\alpha)+ \sqrt{v^2(1+\alpha)^2-4\alpha}}{2}|$  I know that using triangle inequality method $|Z_1|$ is:  $|Z_1|= |\frac{v(1+\alpha)}{2}| + |\frac{\sqrt{v^2(1+\alpha)^2-4\alpha}}{2}|$  Case I: $-1 \leq \alpha \leq 0$ and  $0 &lt; v &lt;1$   Prove that $|Z_1| \leq 1$  Triangle inequality |x+y|=|x|+|y|  I've been stuck on this problem now for a couple of days and I'm having a difficult time proving this case. I was wondering if anybody can assistance me on this problem. I want to thank you ahead of time for your cooperation.
23115
$(i)$ $a&gt;0,b&gt;0,p&gt;0,q&gt;0$   $(ii)$ $\frac{1}{p}+\frac{1}{q}=1$     How can we prove it?   ${a}{b}≤\frac{a^p}{p}+\frac{b^q}{q}$
72090
Show that $z\bar{a}+\bar{z}a \leq 2|a||z|$  I was able to check this using some examples which is not ideal for a mathematician way of proving a problem/case/theorem. I couldn't generalize it (prove it). I will be glad if I can be given the hint/proof for this problem. Thank you.
20426
The entire exercise was like $$8xyz≤(x+y)(y+z)(x+z)≤\frac8{27}(x+y+z)^3\quad x,y,z&gt;0$$  and I managed to solve (demonstrate) the first inequality (from the left) using $$\frac{x+y}2≥\sqrt{xy}$$ but I can't solve the second one. I tried several ideas but I haven't succeeded. Thanks a lot.
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.
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 $\frac{x+y}{2}ln(x) + \frac{x+y}{2} ln(y) \geq xln(\frac{x+y}{2})+yln(\frac{x+y}{2})$
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 :)
22888
Prove that in any triangle $ABC$ the following inequality holds $$\cos{A}+\cos{B}+\cos{C}\ge\dfrac{1}{4}(3+\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})$$  And I have gotten $$8(\cos{A}+\cos{B}+\cos{C})\ge 6+2(\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})$$  $$2(\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})+3=(\sum_{cyc}\cos{A})^2+(\sum_{cyc}\sin{A})^2$$ $$\Longleftrightarrow 8\sum_{cyc}\cos{A}\ge 3+(\sum_{cyc}\cos{A})^2+(\sum_{cyc}\sin{A})^2$$ then Any hints, ideas? Thanks in advance.
20432
Give $a, b. c$ be the lengths of a triangle and $a+ b+ c= 3$. Prove that: $$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a}\geq  \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}$$ My try  We have England MO inequality: $$\left ( \frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \right )^{2}\geq  \left ( a+ b+ c \right )\left ( \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c} \right )$$  and $$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \geq  3=  a+ b+ c$$  But it cannot be used to prove the above inequality and how to use triangle inequality in this inequality. I need the help. Thanks!
48882
Towards the end of , how do I get  $$ \frac{\sqrt{8z + 1} - 1}{2} &lt; w + 1 $$  Here $w = x + y \geq 0, t = (w^2 + w)/2 , z = t + w, x,y \in \mathbb N_{\geq 0}$.  Thanks.
28581
Question: Prove that if $x$, $y$, $z$ are positive real numbers then the following inequality holds: $$\frac{x+y}{x^2+y^2}+\frac{y+z}{y^2+z^2}+\frac{z+x}{z^2+x^2}\leq\frac{1}{x}+\frac{1}{y}+\frac{1}{z}.$$I tried thinking of applying QM-AM, but I didn't know what variables to use. I also thought that this could be related to Cauchy's inequality, but I wasn't sure, again, of what variables to use. Any help would be greatly appreciated, thanks!
82159
let $x,y,z$ are postive numbers,show that $$\dfrac{x+y}{\sqrt{x^2+xy+y^2+yz}}+\dfrac{y+z}{\sqrt{y^2+yz+z^2+zx}}+\dfrac{z+x}{\sqrt{z^2+zx+x^2+xy}}\ge 2+\sqrt{\dfrac{xy+yz+xz}{x^2+y^2+z^2}}$$  My try: Without loss of generality，we assume that $$x+y+z=1$$ and use Holder inequality,we have $$\left(\sum_{cyc}\dfrac{x+y}{\sqrt{x^2+xy+y^2+yz}}\right)^2\left(\sum_{cyc}((x+y)(x^2+xy+y^2+yz)\right)\ge\left(\sum_{cyc}(x+y)\right)^3$$ then we only prove this $$\dfrac{8}{\sum_{cyc}(x+y)(x^2+xy+y^2+yz)}\ge  2+\sqrt{\dfrac{xy+yz+xz}{x^2+y^2+z^2}}$$ then I can't  Thank you
101605
I am trying to prove the above statement but I'm not sure if my proof is correct. My proof is as follows,  Given $u\cdot v$, we know by the C-E Inequality that $|u \cdot v| \leq ||u|| \ ||v||$  Consider $\frac{1}{4}||u+v||^2 - \frac{1}{4}||u-v||^2$  We can manipulate this as follows $$\frac{1}{4}(||u+v||^2 - ||u-v||^2) = \frac{1}{4}(||u+v||- ||u-v||)(||u+v||+||u-v||)$$ Using the triangle inequality the above can be written as follows, $$...\leq \frac{1}{4}(||u|| +||v|| -||u||+||v||) (||u|| +||v|| +||u||-||v||) = \frac{1}{4} (4||u||||v||) = ||u||||v||$$  As such both $u\cdot v$ and $u\cdot v = \frac{1}{4}||u+v||^2 - \frac{1}{4}||u-v||^2 \forall u,v \in \mathbb{R^n}$ are $\leq ||u|| ||v||$  I'm not sure whether I've proven this correctly because I ended up proving that both sides of the equation are less than or equal to something entirely different (the magnitude of v multiplied by the magnitude of u). Is this acceptable? I don't think it is, since if both can be less than or equal to the magnitudes multiplied then they do not necessarily have to be equal to each other. If this is the case, can someone give me a hint on how to move forward with this proof?
106634
Let $G$ be the interior of the circle in the complex plane given by $|z-(1+i)|=1$. Prove that if $z\in G$ then $\sqrt{5}-1&lt;|z-3|&lt;\sqrt{5}+1$. Attempt so far: We know by the triangle inequality that $|a+b|\leq |a|+|b|$ and also $|a-b|\geq||a|-|b||$. Now, $G$ is a circle centred at $(1+i)$ with radius $1$. Using the inequality above, we have $|z-3|\leq |z|+3$, and so $|z-3|\leq 4$. For the lower bound, $|z-3|\geq ||z|-3|=|\sqrt{2}-3|$. This isn't the required inequality though. Any ideas?
15774
If $x&gt;0$, $y&gt;0$, $z&gt;0$ and $xyz = 1$ then   $$\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2} \le \sqrt{2}(x+y+z)$$   I tried using $\displaystyle x = \frac{a}{b},y = \frac{b}{c}$ and $\displaystyle z = \frac{c}{a}$ substitution,  $\displaystyle \sum_{cyc} \frac{\sqrt{a^2+b^2}}{b} \le \sqrt{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) \\ \displaystyle  \iff 6+ 2\sum_{cyc}\frac{\sqrt{(a^2+b^2)(b^2+c^2)}}{bc} \le \sum_{cyc}\frac{a^2+b^2}{b^2} + 4\sum_{cyc}\frac{b}{a}$  So it would suffice if we showed $\displaystyle \sum_{cyc}\frac{\sqrt{(a^2+b^2)(b^2+c^2)}}{bc} \le 2\sum_{cyc}\frac{b}{a}$  Squaring again does not lead anywhere nice. Is the last inequality true ? If not is there a better way of showing the result ?
106384
An inequality proposed at Zhautykov Olympiad 2008.     Let be $a,b,c &gt;0$ with $abc=1$. Prove that:    $$\sum_{\mathrm{cyc}}{\frac{1}{(a+b)b}} \geq \frac{3}{2}.$$   Set $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$.   Our inequality becomes: $$\sum_{\mathrm{cyc}}{\frac{z^2}{zx+y^2}} \geq \frac{3}{2}.$$  Now we use that: $z^2+x^2 \geq 2zx.$ $$\sum_{\mathrm{cyc}}{\frac{z^2}{zx+y^2}} \geq \sum_{\mathrm{cyc}}{\frac{2z^2}{z^2+x^2+2y^2}} \geq \frac{3}{2}.$$  Now applying Cauchy-Schwarz we obtain the desired result.  What I wrote can be found on this link: . But now, I don't know how to apply Cauchy-Schwarz.   Thanks:)
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.