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.
such that 
, where 
are non-negative reals such that 
.
.
a tangencial quadrilateral that is not a rhombus, let 
be lengths of tangents from 
to the incircle of the quadrilateral which center is 
. Let 
be the midpoints of 
resp. Prove that![\[ \frac{MI}{IN}=\frac{a+c}{b+d} \]](http://latex.artofproblemsolving.com/e/7/0/e709acb8b17c0466a385ba8e7a89bb1bf16c72af.png)
is given. It turns out that for every positive integer number 
, for which the numbers 
and 
are positive, they have the same number of divisors. Is it necessarily true that 
?
with as its incenter. Let 
be the intersection of 
and 
and define 
in a similar way. Furthermore, let 
. Prove that if 
, then 
are concyclic.
positive integers. Product of any one of them with sum of remaining numbers increased by 
is divisible with sum of all numbers. Prove that sum of squares of all numbers is divisible with sum of all numbers
![$P \in \mathbb R[x,y]$](http://latex.artofproblemsolving.com/8/a/0/8a0576299e976c51607959fb68c7c742b2a49c6a.png)
with ![$P \not\in \mathbb R[x] \cup \mathbb R[y]$](http://latex.artofproblemsolving.com/a/f/7/af71ba8c643d5bd8a2928fa6714ccd91713bd641.png)
and 
homeomorphismes of 
, 
is an homoemorphisme too.
is inscribed in a circle centered at 
, and 
. Let 
be the second intersection point of the circumcircles of the triangles 
and 
.
, and are collinear.
and 
are the circumcenters of the triangles , and , respectively, prove that the points 
are concyclic.
where is a positive integer and 
is a prime.(Don't get mad at me,I've used the search function and did not see a correct and complete solution anywhere.)
![\[
2^a + 3^b + 1 = 6^c
\]](http://latex.artofproblemsolving.com/1/9/e/19eb5efa2ccdf54cd277339f8d3b99f75168779d.png)
is inscribed in circle 
. Let 
and denote its orthocenter and circumcenter, respectively. Let 
and 
be the midpoints of sides 
and 
, respectively. Rays 
and 
meet at 
and 
, respectively. Lines 
and 
meet at . Prove that 
.
satisfying both of the following properties:
.
.
and 
for each nonnegative integer 
. Prove that 
is the greatest integer less than or equal to 
, 
and 
equals one, then 
not 
But with the same substitutions, we can use alternative C-S solution (and Vasile's inequality) like below:
not But with the same substitutions, we can use alternative C-S solution (and Vasile's inequality) like below:
and be positive numbers . Prove that 
be positive numbers . Prove that 
and be positive numbers . Prove that 
, the Cauchy-Schwarz's Inequality gives that
It is easy to show that 
. Therefore, the inequality ![$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{3}{\sqrt[3]{abc}}$](http://latex.artofproblemsolving.com/f/4/1/f41b3d1ee10a3656bbae0ce631a1c38b24d6b96c.png)
implies![\begin{align*}\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge\frac{\frac{9}{\sqrt[3]{(abc)^2}}}{3+\frac{3}{\sqrt[3]{abc}}}=\frac{3}{\sqrt[3]{abc}(\sqrt[3]{abc}+1)}\ge\frac{3}{1+abc},
\end{align*}](http://latex.artofproblemsolving.com/8/4/b/84b96b1dabde2948d23d6ab6fcd9e2ac53f94973.png)
where the last inequality holds since 
. We are done. 
. Prove that
(Lijvzhi)
Prove that![\[ \frac 1{a(2+a)}+\frac 1{b(2+b)}+\frac 1{c(2+c)}+\frac 1{d(2+d)}\geq \frac 4{ 2+abcd}\]](http://latex.artofproblemsolving.com/0/4/d/04d223126d59422752a97e762b601cc298191683.png)
Solution of ytChen 
. Using the Cauchy-Schwarz's Inequality and AM-GM Inequality, we obtain![\begin{align*}\frac 1{a(2+a)}+\frac 1{b(2+b)}+\frac 1{c(2+c)}+\frac 1{d(2+d)}=&\frac{x^2}{2x+1}+\frac{y^2}{2y+1}+\frac{z^2}{2z+1}+\frac{u^2}{2u+1}\\
\ge&\frac{(x+y+z+u)^2}{2(x+y+z+u)+4}\\
=&\frac{1}{\frac{2}{x+y+z+u}+\frac{4}{(x+y+z+u)^2}}\\
\ge&\frac{1}{\frac{1}{2\sqrt[4]{xyzu}}+\frac{1}{4\sqrt[4]{(xyzu)^2}}}\\
=&\frac{4}{2\sqrt[4]{abcd}+\sqrt[4]{(abcd)^2}}\\
\ge&\frac{4}{2+abcd},
\end{align*}](http://latex.artofproblemsolving.com/3/e/5/3e5b9b4d313d736f137b85c77060bd4cc4455000.png)
where the last inequality holds since ![$w^4+2-w^2-2w=(w-1)^2(w^2+2w+2)\ge0\ \left(\forall w=\sqrt[4]{abcd}\right)$](http://latex.artofproblemsolving.com/d/3/b/d3b87e901d52d2a6b3ba7dd5104500674c68b4ed.png)
. We are done.
. Prove that(Lijvzhi)
. Using the C-S and AM-GM , we obtain![\begin{align*}\frac{1}{a_1\left(a_1+n-2\right)}+\frac{1}{a_2\left(a_2+n-2\right)}+\cdots+\frac{1}{a_n\left(a_n+n-2\right)}\\=&\frac{b_1^2}{(n-2)b_1+1}+\frac{b_2^2}{(n-2)b_2+1}+\cdots+\frac{b_n^2}{(n-2)b_n+1}\\
\ge&\frac{(b_1+b_2+\cdots+b_n)^2}{(n-2)(b_1+b_2+\cdots+b_n)+n}\\
=&\frac{1}{\frac{n-2}{b_1+b_2+\cdots+b_n}+\frac{n}{(b_1+b_2+\cdots+b_n)^2}}\\
\ge&\frac{1}{\frac{n-2}{n\sqrt[n]{b_1b_2\cdots b_n}}+\frac{1}{n\sqrt[n]{(b_1b_2\cdots b_n)^2}}}\\
=&\frac{n}{(n-2)\sqrt[n]{a_1a_2\cdots a_n}+\sqrt[n]{(a_1a_2\cdots a_n)^2}}\\
\ge&\frac{n}{n-2+a_1a_2\cdots a_n},
\end{align*}](http://latex.artofproblemsolving.com/8/1/2/81201ba705423fef3914cd56b28fac342bfc815a.png)
where the last inequality holds since 
![$\forall w=\sqrt[n]{a_1a_2\cdots a_n}$](http://latex.artofproblemsolving.com/a/b/f/abf9b837da0fb966837e253cac8a8ebe7fc0e365.png)
. We are done. 
. Prove that
Let 
. Prove that
. Prove that