Consider three real numbers a,b,c such that $a^2+b^2+c^2 =1$. What will be the maximum value of the expression $|a-b| + |b-c| + |c-a|$ ? I tried two approaches. One I used the fact that $a^2+b^2+c^2 =1 = \frac{1}{2} (|a-b|^2 + |b-c|^2 + |c-a|^2) + ab + bc + ca$. Original plan was to somehow eliminate ab + bc + ca , but unable to do so - got stuck in a series of repeating expressions Another variant was to try and assert that $a \geq b \geq c$ without loss of generality. Then $(|a-b| + |b-c| + |c-a|)^2 \leq 4(a-c)^2$ , but I am unable to use the given fact about $a^2 + b^2 + c^2$. I realize that I am missing something very trivial here as this seems a fairly simple problem - so any help would be greatly appreciated.
81424
Given reals $a_1, a_2, \cdots, a_{n - 1}, a_n$ such that $a_1^2 + a_2^2 + \cdots + a_{n - 1}^2 + a_n^2 = 1$ $(n \in \mathbb N, n \ge 3)$. Calculate the maximum value of $$\large |a_1 - a_2| + |a_2 - a_3| + \cdots + |a_{n - 1} - a_n| + |a_n - a_1|$$   There must exist $1 &lt; k &lt; n$ such that $a_{k - 1} \le a_k \le a_{k + 1}$  $ \implies |a_{k + 1} - a_k| + |a_k - a_{k + 1}| = |a_{k + 1} - a_{k + 1}|$.  Repeat the above process for about $n - 1$ times and we have that $$|a_1 - a_2| + |a_2 - a_3| + \cdots + |a_{n - 1} - a_n| + |a_n - a_1| \le 2 \cdot \min(|a_i - a_j|, 1 \le i &lt; j \le n)$$  Now we just have to find the maximum value of $\min(|a_i - a_j|, 1 \le i &lt; j \le n)$ for $$a_1^2 + a_2^2 + \cdots + a_{n - 1}^2 + a_n^2 = 1$$, which I don't know how.
764
I am stuck on these two problems.   $1$. Prove that for every three positive real numbers a, b, and c that   $(a+b+c)*(\frac{1}{a}+\frac{1}{b} + \frac{1}{c}) \ge 9$.   $2$. Prove that for every three positive real numbers a, b, and c that  $a^2 + b^2 + c^2  \ge ab + bc + ac$.  I have tried direct proof and have not gotten anywhere significant. I won't put the work on there since it is way too long and I don't think it will help. There must be some sort of trick involved, but for the life of me, I cannot figure it out.
122667
What is the maximum value of $x^2+y^2$, where $(x,y)$ are solutions to $2x^2+5xy+3y^2=2$ and $6x^2+8xy+4y^2=3$. (calculus is not allowed). I tried everything I could but whenever I got for example $or$ $x^2+y^2=f(y)$ or $f(x)$ the function $f$ would always be a concave up parabola, so I could not find a maximum for either variable. However, I also don't see how you could solve it if you leave both variables on one side. And by the way I know that you can solve for $x$ and $y$ using the quadratic formula and get $4$ different solutions but I am looking for a much more efficient way than that.   This question came from a math competition from the Math Honor Society, Mu Alpha Theta.
25531
$a,b,c$ are positive reals with $abc = 1$. Prove that    $$\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$$   I try to use AM $\ge$ HM.  $$\frac{\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(a+b)}}3\ge \frac{3}{a^3(b+c)+b^3(a+c)+c^3(a+b)}$$ Then how I proceed.
74202
I have been working on the following problem:     For a given $c\in \Bbb Z^+$, find $a,b \in \Bbb Z^+: c=a^2-b^2$.   I have already figured out and proved a number of things:   A way to directly determine if a particular instance is solvable:  $\exists a,b \in \Bbb Z^+: c=a^2-b^2 \iff c \not\equiv 2 \mod 4$ An upper bound on the values for a (and b): $b \lt a\le\lfloor\frac{c-1}{2}\rfloor$ And a simple algorithm to find a solution:     $$ \begin{align*}  (a_0, b_0) &amp;= (0, 0) \\  (a_{n+1}, b_{n+1}) &amp;= \begin{cases}   (a_n+1, &amp;b_n&amp;)&amp;\text{if } a_n^2-b_n^2 &lt; c\\   (a_n, &amp;b_n+1&amp;)&amp;\text{if } a_n^2-b_n^2 &gt; c\\  \end{cases} \end{align*} $$   This can be iterated until a solution is found or the upper bound is reached.   I am currently trying to work out a more direct way to find an answer, but I have been unable to figure out any other ways to approach the problem.  How can I more directly find values for $a$ and $b$ for a given value of $c$?
31427
If a,b,c are real numbers satisfying $a+b+c =0; a^2+b^2+c^2=1$.  Prove that $a^2 b^2 c^2 \le \frac{1}{54}$.
50245
Suppose that $a,b,c$ are the lengths of the sides of a triangle. Prove that $$a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc.$$   Attempt  It's hard to do much simplification here so we are going to have to use that $a &gt; b+c, b&gt;a+b, c&gt;a+c.$ Using that fact we obtain that $$b^2(c-b)+c^2(b-c) &gt; a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c).$$ Then should I use AM-GM to finish it or what inequality?
12722
If $a,b,c,d$ are positive integers with    $$a+b+c+d=63,$$   what is the maximum value of $$ab+bc+cd?$$   I treated this as an optimization problem and tried to solve it but I did not make any progress.
61349
Let $a$, $b$, and $c$ be real numbers such that $a − 7b + 8c = 4$ and $8a + 4b − c = 7$. What is the value of $a^2-b^2+c^2$ ?
21345
Find the maximum value of $$(1-x)(2-y)(3-z)(x+\frac y2+\frac z3)$$ where $$x&lt;1,y&lt;2,z&lt;3,x+\frac y2+\frac z3&gt;0$$  I tried several cases to find a certain product. I tried 7 such cases. Three of them yield negative values while 4 gave postive fractional values. I chose the highest value which is 5/3. I already gave up and looked at the answer at the back of the questionnaire. The answer is $$\frac {243}{128}$$  Can anyone help me in answering this question?
79834
if $x+2y+z=4$ and $x,y,z$ are real number. then find maximum value of $xy+yz+zx$  putting $x+z=4-2z$ in $y(x+z)+zx = y(4-2z)+zx = 4y-2yz+zx$  i wan,t be able to  go further,could some help me with this
40775
a,b,c,d $\in \mathbb{R}$  $a,b,c,d \gt 0$ and  $ c^2 +d^2=(a^2 +b^2)^3$ prove that  $$ \frac{a^3}{c} + \frac{b^3}{d} \ge 1$$  If I rewrite the inequation like $ \frac{a^3}{c} + \frac{b^3}{d} \ge \frac{c^2 +d^2}{(a^2 +b^2)^3}$ and manage to simplfy  it brings me nowhere. I try with  Cauchy-Schwarz Inequality but still can not solve it. I would very much appreciate it if anyone could help me
115575
If $a$, $b$, $c$, $d$ belong to $\mathbb{R}$, and   $$(a+b)(c+d)=143 \qquad (a+c)(b+d)=150 \qquad (a+d)(b+c)=169$$   Find the smallest possible value of    $$a^2 + b^2 + c^2+ d^2$$   I thought of adding $7$ to the first equation and make it equal the second, solving, and finding a new equation.  Do it $3$ times, keep substituting, but this is a very very very long approach if it's even an approach.
91562
$a,b,c$ are positive real numbers such that $a+b+c=1$. Prove that: $$\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2} \geqslant \dfrac{1}{2} $$ I have tried with Cauchy-Schwarz inequality in Engel form...
24300
Let $a, b, c, d, e, f$ be nonnegative real numbers such that   $a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6$. $ab + cd + ef = 3$. What is the maximum value of $a+b+c+d+e+f$&nbsp;?.   I was thinking about somehow manipulating the numbers to get $\left(a + b + c + d + e + f\right)^{2}$, but I don't really know what to do.
44952
Let $a$, $b$, and $c$ be positive real numbers. Prove that  $$\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}$$  Under what conditions does equality occur? That is, for what values of $a$, $b$, and $c$ are the two sides equal?
33054
I have a  beautiful  inequality, but I can only prove part of cases. Given any positive real numbers $a,b,c$, we have   $$(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$$  How can we prove this statement?