where 
is a set of matrix size 
with complex entries. Let 
be polynomial characteristic of 
and ![$q(x) \in \mathbb{C}[x]$](http://latex.artofproblemsolving.com/2/b/a/2ba4216f2017e1e92484e9c69cb1ae87d5ea723b.png)
is monic polynomial such that 
and 
.
?
be reals such that 
Prove that
and 
be independent normally distributed random variables, each with its own mean and variance. Show that the variance of 
is smaller than the variance of 
be an integer and let 
be non-negative real numbers. Prove that:![\[
(x_1 + x_2 + \dots + x_n)^2 \geq 4(x_1x_2 + x_2x_3 + \dots + x_{n-1}x_n + x_nx_1)
\]](http://latex.artofproblemsolving.com/5/4/b/54bd17c29d2397a13e1cd377bccd324e7bff74a6.png)
and 
.
Let 
. Prove that
![$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{3\sqrt[3]{3}(a+b+c)}{\sqrt[3]{4}(1+3abc)}.$$](http://latex.artofproblemsolving.com/b/e/0/be03dc2775c794bef7f002c7a999c7372fa517eb.png)
such that
and 
are one-third of their respective sides. It follows that 
,
and 
Then the area of triangle 
is:
such that ![$\sqrt[3]{1}=x$](http://latex.artofproblemsolving.com/e/d/9/ed965489f2f983dbb4669c9e94ea9aa83baa72a7.png)
.
such that ![$\sqrt[4]{1}=y$](http://latex.artofproblemsolving.com/d/1/6/d16559d02217f335c54309bd3effb04e6e2f4b01.png)
.
such that ![$\sqrt[5]{1}=z$](http://latex.artofproblemsolving.com/3/3/d/33d87611165181f89f950addf1afb749c1b01df9.png)
.
.
are the roots of the equation 
.
.
Y by Adventure10, Mango247
For 
a list of 
real numbers 
is said to be area definite for 
if the inequality ![\[\sum_{1\le i<j<k\le m}a_{ijk}\cdot\text{Area}(\triangle A_iA_jA_k)\ge0\]](//latex.artofproblemsolving.com/e/7/7/e778445aa49547e4a227d16cc8685370a5d1a7d9.png)
holds for every choice of 
points 
in 
For example, the list of four numbers 
is area definite for 
Prove that if a list of numbers is area definite for 
then it is area definite for 
a list of 
real numbers 
is said to be area definite for 
if the inequality ![\[\sum_{1\le i<j<k\le m}a_{ijk}\cdot\text{Area}(\triangle A_iA_jA_k)\ge0\]](http://latex.artofproblemsolving.com/e/7/7/e778445aa49547e4a227d16cc8685370a5d1a7d9.png)
holds for every choice of 
points 
in 
For example, the list of four numbers 
is area definite for 
Prove that if a list of numbers is area definite for 
then it is area definite for 
multiplies the area of a triangle by some constant 
on average. (We do not need the value, just that by symmetry it is the same for all triangles.) Letting 
be the sum and 
the sum after the random projection, we have 
.
then 
and so is its expectation.
for a triangle 
is a random projection independent of 
.
L
be the projections of the 
- which we will denote 
- is non-negative for all planes 
,
,
,
to 
be a unit normal vector to the plane of 
,![\[\mathrm{area}(\bigtriangleup B_iB_jB_k) = |\vec{n}\cdot\vec{v}_{ijk}|\;\mathrm{area}(\bigtriangleup A_iA_jA_k).\]](http://latex.artofproblemsolving.com/0/f/a/0fa64a7aff055554ce211af38b6a4597d81ccaa7.png)
We sum over all 
with weights ![\[\int_{S^2} \Sigma(P) = \sum_{i,j,k} a_{ijk}\;\mathrm{area}(\bigtriangleup A_iA_jA_k) \int_{S^2} |\vec{n}\cdot\vec{v}_{ijk}|.\]](http://latex.artofproblemsolving.com/4/d/2/4d249e04b374566d51155bc9e2e919cfb6da48f2.png)
What remo is arguing is that 
is some absolute positive constant 
independent of ![\[\sum_{i,j,k} a_{ijk}\;\mathrm{area}(A_iA_jA_k) = \frac{1}{C}\int_{S^2} \Sigma(P) \ge 0.\]](http://latex.artofproblemsolving.com/b/6/6/b661e1b1075a8bb038ab70f70952d85452da52ba.png)
So 
as desired.
instead of 
-
as it moves over unit vectors.
when projected onto a plane with normal vector 
be 
and 
.![$[\triangle A_iA_jA_k]=\frac{1}{2}|\vec{u}\times \vec{v}|$](http://latex.artofproblemsolving.com/f/9/3/f93836035b3e6eb87eddbe3f974b46711c3538a6.png)
.
and hence the projection of 
.
A quick expansion using the identity 
gives that the projected area is
Where 
is the angle between the plane containing 
and the plane we're projecting onto. Switching from unsigned area to signed area this becomes 
For a fixed choice of 
we integrate over planes in every direction (specifically, tangency planes to the unit sphere at every point):![$$\iint\limits_{\text{unit sphere}}[\triangle A_iA_jA_k]|\cos\theta|\mathrm{d}A=[\triangle A_iA_jA_k]\iint\limits_{\text{unit sphere}}|\cos\theta|\mathrm{d}A$$](http://latex.artofproblemsolving.com/1/0/c/10ce7a9ef2d6c9b4fad6c9242b0c093f918b0a89.png)
Where 
is the angle between the plane tangent to the piece of area 
on the unit sphere and the plane of 
.
,
Therefore, we have that![$$\sum_{i,j,k}a_{ijk}[\triangle A_i A_j A_k]=\sum_{i,j,k}a_{i,j,k}\frac{\iint_{\text{unit sphere}}A_{\text{projected}}\mathrm{d}A}{C}$$](http://latex.artofproblemsolving.com/1/4/6/146150e6caf1f4b1f60ab3e64bfd3c28f0a8083c.png)
Because 
is area definite for each individual plane.
.
with finite area 
and randomly orient it in space and project onto a fixed plane, then the expected area of the projection is 
.
in space, then project it onto a fixed plane yielding a set 
then![\[ 0 \le \mathbb{E} \left [ \sum_{i < j < k} a_{ijk} \cdot \text{Area}(\triangle A_i'A_j'A_k') \right ] =
\sum_{i < j < k} a_{ijk} \cdot \mathbb{E} [ \text{Area}(\triangle A_i'A_j'A_k') ] = \frac 12 \sum_{i<j<k} a_{ijk} \cdot \text{Area}(\triangle A_iA_jA_k) \]](http://latex.artofproblemsolving.com/3/b/1/3b14cbc277e91e6d01c1fd960ffe4eb7f6d4b9e3.png)
![\[ \implies \sum_{i<j<k} a_{ijk} \cdot \text{Area}(\triangle A_iA_jA_k) \ge 0 \]](http://latex.artofproblemsolving.com/9/b/0/9b0ff7e8409d3809227688b3e140f70c2e65cbf2.png)
implying its area definite.
