If $x,y,z$ are distinct positive numbers, such that $$x + \frac1y = y + \frac1z = z + \frac1x $$ then value of $xyz$ is?  $$A)\ 4\quad  B)\ 3\quad C)\ 2\quad  D)\ 1$$  My attempt:  1.I equaled the equation to '$k$'. Using the AM-GM inequality, I found that $k&gt;2$ (the equality does not hold because all are distinct). However, this, I couldn't put to much use. 2. For the next attempt, I substituted the values of $x$ and $y$ in terms of $z$ and $k$ in $xyz$. What I am getting is $xyz=zk^2 - k - z$. That's the farthest I could do..Please, help.
6568
If $a, b, c$ be distinct reals such that $$a + \frac1b = b + \frac1c = c + \frac1a$$ how do I find the value of $abc$?  The answer says $1$, but I am not sure how to derive it.
4916
Question:  let $x,y,z$ be postive integers and  $x\neq y\neq z$ such $$xyz|(xy-1)(xz-1)(yz-1)$$ Find all $x,y,z$,  my try: $$(xy-1)(yz-1)(xz-1)=x^2y^2z^2-xyz(x+y+z)+(xy+yz+xz)-1$$ then  $$\dfrac{(xy-1)(yz-1)(xz-1)}{xyz}=xyz-(x+y+z)+\dfrac{xy+zy+xz-1}{xyz}$$ so $$\dfrac{xy+yz+xz-1}{xyz}\in N$$ Then I can't get any further  I found $(x,y,z)=(2,3,5),(2,5,3),(3,2,5),(3,5,2),(5,2,3),(5,3,2)$such it, I don't know/have any other,  also How to prove it.  Thank you very much
44581
Given that $a,b,c$ are distinct positive real numbers, prove that $(a + b +c)\big( \frac1{a}+ \frac1{b} + \frac1{c}\big)&gt;9$   This is how I tried doing it:  Let $p= a + b + c,$ and $q=\frac1{a}+ \frac1{b} + \frac1{c}$.  Using AM>GM for $p, q$, I get:  $$\frac{p+q}{2} &gt; {(pq)}^{1/2}$$ $$\sqrt{(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)} &lt; \frac{\big(a+\frac1{a} + b+\frac1{b} + c+\frac1{c}\big)}2$$ And for any $x\in \mathbb{R}, \space \space x+\frac1{x}≥2.$ Thus,   $$(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)&lt;9, $$ which is the opposite of what had to be proven. What did I do wrong?
14943
If $a, b$ are positive real numbers and $a+b = 1$, prove that :  $$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}$$  I can see that the value $\frac{25}2$ is attained for $a=b=\frac12$. But I do not know how to show that this is the minimal possible value.  Thank you.
11884
Write down the set of distinct solutions and prove your list is complete. $x$ and $y$ are positive integers.  I have rewritten it as $\frac{(x+y)}{xy} = \frac1p$, but I don't understand where to go from here?
18735
Prove that, $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}\ge 4$ where we do not use AM-GM inequality on the given statement to prove it. Typically, I am actually looking for a little advanced and elegant solution.    EDIT: $a,b,c,d&gt;0$
109144
Given positive real numbers $x,y,z$ satisfying $x+y+z=3$, prove that $$ \sqrt{3x+\frac{1}{y}}+\sqrt{3y+\frac{1}{z}}+\sqrt{3z+\frac{1}{x}}\geqslant 6. $$
44720
There is a unique triplet of positive integers $(a, b, c)$ such that $a ≤ b ≤ c$.  $$ \frac{25}{84} = \frac{1}{a} + \frac{1}{ab} + \frac{1}{abc} $$  Just having trouble with this Canadian Math Olympiad question. My thought process going into this, is:  Could we solve for $\frac{1}{a}$ in terms of the other variables? Then substitute that value in for each occurrence of $a$, to solve for $a$?  That's all I can really think of right now. It's a question I'm not exactly used to... It's sort of the first of these kinds that I've faced.  Thanks.
79847
If $z$ is any non-zero complex number, how to prove the following inequality? $$ | z-1 | \le | | z | -1 | + | z| \cdot | \arg z | $$  Hints please!
71836
Let $a, b, c, d &gt; 0$. Prove that $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq a + b + c + d$.   I'm supposed to prove this by AM-GM, but I can't see how. Any help would be appreciated.
61045
This is Problem 6 of the 2007 Indian National Math Olympiad (INMO).  If $x, y, z$ are positive real numbers, prove that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2).$  My best idea was to expand this and simplify. Although that doesn't look very feasible. Another idea is to see that $x^2+y^2+xy \geq x^2+y^2$. Then we just have to show that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(x^2+y^2)(x^2+z^2)(y^2+z^2)$ if that is even true.
20411
Question:     Given complex numbers $a,b,c$, we have that $|az^2 + bz +c| \leq 1$ holds true for any complex number $z, |z| \leq 1$. Find the maximum value of $|bc|$   It is said this is answer is $$|bc|\le \dfrac{3\sqrt{3}}{16}$$  My idea: let $z=1$,then $$|a+b+c|\le 1$$ let $z=-1$,then $$|a-b+c|\le 1$$ let $z=0$, then $$|c|\le 1$$  let $|\theta|=1$,then we have $$|a(z/\theta)^2+b(z/\theta)+c|\le 1\Longrightarrow |az^2+b\theta z+c\theta^2|\le 1$$ and only this can't solve this problem,Thank you
77585
if $x,y,z$ are positive real numbers,Prove:$$3(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\ge xyz(x+y+z)^3$$ Additional info:$\sum_{cyc}$ denotes sums over cyclic permutations of the symbols $x,y,z$. I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.  Things I have done: Writing Left hand side in expanded form:$$LHS=3(\sum_{cyc}x^3y^3+\sum_{cyc}xy^4z+3x^2y^2z^2)$$ Right now I don't have good idea were to start(Is expanding and looking for AM-GM is good?).So any hint is appreciated. Source : This inequality can be found as Problem 42, page 13, in the book &quot;Old and New inequalities&quot; by Titu Andreescu, where it is attributed to Manlio Marangelli.
113922
I can't see any ways I could use the fact that $a+b+c=1$. I've tried solving the inequality using various AM-GM inequalities, but I just can't make it. I need some help. Thanks.
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.
58414
I want to prove the following inequality :      $$\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab,  $$   $a,b,c,d \in \mathbb{R}_{+} .$    In my book, at the answers chapter the author uses AM $\geq$ GM, but I haven't any idea how I can use that.   Thanks :)
101231
Let, $x,y,z$ be consecutive positive integers such that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} &gt; \frac{1}{10}$ I need to find the maximum value of $x+y+z$ ?  My attempt:  I tried guessing first, To maximise $x+y+z$ it is necessary to have high value for $x$, But not too high which will contradict the constrain of the problem. As we can roughly split the $0.1$ into three parts so I started with numbers closer to $30$ with trial an error I got that for $x=29$ we get the optimum and final result would be 90 (=max. value of $x+y+z$).   Is there any way to arrive at this conclusion with using inequalities? Any hints will be appreciated...