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.
. [/center]
, and 
, 
. This is proven in solution 1a-1. We will refer to this as Trick 1.
prove (or disprove) that there exists infinitely many positive integers n in which sums to k! where k is a positive integer. (examples, 1+2+3=3! and 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15=5!
and 
be two positive five-digit palindrome integers, such that 
and there is no other palindrome number between them. Determine the possible values of 
.
to 
. Vlad has 
coins, the sum of denominations of which is 
. Find the minimal possible value of 
at which we can confidently say that Vlad is able to pay without change.
, let 
be the largest prime number that divides 
. For 
, define 
. such that the numbers 
, 
, and 
are all less than 
., prove that there exists a positive integer such that the numbers , , and are all less than 
.
be a cyclic quadrilateral. Let 
be a point on the ray 
which is outside the circumscribed circumference of . Suppose that 
. Prove that 
if and only if 
.
. Determine all -tuples 
of real numbers which satisfy the equality![\[ b_1^2 + b_2^2 + \cdots + b_k^2 = 4\left(a_1^2 + a_2^2 + \cdots +a_k^2\right)\]](http://latex.artofproblemsolving.com/1/f/8/1f8225dd21e6249fff12f0c072daf9355b3ac927.png)
where 
for 
.
with 
such that 
for every 
.
and be points on a circle in this order. The chords 
and 
intersect in point 
. The perpendicular to through C and the perpendicular to through intersect in point 
.
and 
are perpendicular.
, and t is the tangent to the circumcircle of ABC at A. Finally, let E be the intersection of and t. Show that CE and t are perpendicular to each other.
be non-negative real numbers. Prove that![\[\frac{3(a^2 + b^2 + c^2 + d^2)}{ab + bc + ca + da + db + dc} + \frac{256abcd}{(a + b + c + d)^4} \geqslant 3.\]](http://latex.artofproblemsolving.com/0/5/d/05dc58b0b3bbc1347ed7fa2a837b8f197b2c17b1.png)
, where 
always occurs at 
, where 
. We want to prove that its minimum value for 
occurs at 
. with respect to 
:
to find critical points:
Since 
, 
. We can factor out 
(which is non-zero):
As 
, we must have:
Let 
. Since 
, 
. Thus 
, which implies 
. This critical point is in the domain .
., 
. The sign of 
is thus determined by the sign of the term 
.
, then 
. This implies 
, so 
. Thus, is decreasing on the interval 
.
, then 
. This implies 
, so 
. Thus, is increasing on the interval 
. decreases before 
and increases after , is a local minimum.: at is 
., 
.
, 
.
, 
. is the only critical point in 
and , while the function approaches 
at one end of its domain and 
at the other, the local minimum at is the global minimum. for occurs at .
.
![$p'(x)=(n+1)x^n-nx^{n-1}=x^{n-1}[(n+1)x-n]=0 \Rightarrow x_1=\dfrac{n}{n+1}, x_2=0(\text{not satisfy}) $](http://latex.artofproblemsolving.com/c/8/c/c8c3ecfe7a067ad6f7e5aec087447966add371e8.png)
is a sign reversal zero point of 
which implies it.