AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
, 
, and 
be non-negative real numbers and let 
, 
, and 
be positive real numbers such that 
. Prove that![\[ \dfrac{a^3}{x^2}+\dfrac{b^3}{y^2}+\dfrac{c^3}{z^2} \ge a+b+c.\]](http://latex.artofproblemsolving.com/6/3/4/6343b9d30ed2bbd9c5b81d0a9eb6828505c92b53.png)
is a minuan pair if it satisfies the following two conditions:
is even.
are all the positive divisors of , ordered such that 
, then the set of all positive divisors of 
is precisely
.
![\[ a_n=2^n+3^n+6^n -1,\ n\geq 1. \]](http://latex.artofproblemsolving.com/6/4/9/6499d8f992f3e72765cab1f3afe15df99d01b12a.png)
be positive integers with 
. Find 
.
be an acute-angled triangle with circumcenter 
. Let 
be a circle with center on the altitude from 
in , passing through vertex and points 
and 
on sides 
and 
. Assume that![\[BP\cdot CQ = AP\cdot AQ.\]](http://latex.artofproblemsolving.com/1/c/2/1c22443c125d448bf3f48d2729ae4f2eb375a405.png)
Prove that is tangent to the circumcircle of triangle 
.
number 
is not divisible by .Prove that 
is not divisible by for any natural 
and 
be an odd integer greater than 1 and let 
be integers. For each permutation 
of 
, define 
. Prove that there exist permutations 
of such that 
is a divisor of 
.
be the set of the permutation matrices of dimension . If 
, then 
.
.
![\[ \dfrac {\text {gcd}(m, n)}{n} \dbinom {n}{m} \]](http://latex.artofproblemsolving.com/c/0/a/c0a6877cd49e904728b24e7743fe8021efd6d312.png)
is an integer for all pairs of integers 
.
reaches a global minimum
, meaning one matrix is the inverse of the other. So I tried to factorize 
, which is 
. Hence we can exchange the order and get the other one: 
.
so 
where we use the condition rewritten as 
. Now we are done if we can show 
is invertible because then we can cancel it and get 
which rewrites to 
. Let 
, try some things to see 
and 
, so 
and that means 
is invertible and we are done.
![\[ \begin{bmatrix} C & B \\ 0 & I \end{bmatrix}\begin{bmatrix} I & B \\ 0 & A^{-1} \end{bmatrix} = \begin{bmatrix} C & AC \\ 0 & A^{-1} \end{bmatrix} \]](http://latex.artofproblemsolving.com/6/f/f/6ff37ee00490748c8f67594a3a9bf5f4265c35e1.png)
which clearly follows from the identity given.
we get![\[ \begin{bmatrix} C & B \\ 0 & I \end{bmatrix} = \begin{bmatrix} C & -CBA+ACA \\ 0 & I \end{bmatrix}. \]](http://latex.artofproblemsolving.com/c/c/9/cc9021577b34c0533bc90f356988fdff44c86dc5.png)
Comparing the upper-right entry, we get the desired identity.
for which 
,
be real square matrices of the same order, and suppose ![\[ (A-B)C=BA^{-1}\implies C(A-B)=A^{-1}B \]](http://latex.artofproblemsolving.com/0/6/0/06007271017c2c48faa708366f1adc67db91f245.png)
