and 
Prove that
Let 
and 
Prove that
prove that![\[ \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{d^2} + \frac{d^2}{a^2}\geq\frac{a}{d} + \frac{b}{a} + \frac{c}{b} + \frac{d}{c}\]](http://latex.artofproblemsolving.com/0/0/4/004e7f148bbb0c797688f5195aa4002dbc316470.png)
If you know same tasks, link them please
be the Fermat point of a 
,
,
are concurrent and the point of concurrence is 
,
(
well). i want to study so if you guys have some easy or normal problems please send me
Y by
Let 
have median 
(
) and circumcenter 
. The circumcircle of 
bisects . Determine the least possible perimeter of if it has integer side lengths.
have median 
(
) and circumcenter 
. The circumcircle of 
bisects . Determine the least possible perimeter of if it has integer side lengths.
Let the midpoint of 
.
and 
,
.
,
,
.
and 
,
is a parallelogram, so![\[\angle CAM=\angle CAX+\angle XAM=\angle CBX+\angle CXA=\angle CBX+\angle CBA=\angle ABX=\angle AMN=\angle AMC\Longrightarrow CA=CM\]](http://latex.artofproblemsolving.com/1/b/1/1b1e6219e82593a1e63c27c29bc135abfe66d275.png)
Let 
be the midpoint of 
.
is cyclic, but 
is a midsegment in 
,
and 
is cyclic, therefore 
.
,
,
we can use the median formula 
)![\[b^2=CM^2=\frac{1}{4}(2a^2+2b^2-c^2)\Longrightarrow c^2=2(a-b)(a+b)\]](http://latex.artofproblemsolving.com/8/1/b/81b7181192a00841e3db99b4d4e903cf368aac4d.png)
We want 
to be positive integers with minimal sum (the perimeter of 
,
are the sidelengths of a triangle which satisfies the conditions in the problem with smaller perimeter. Thus we can assume that 
,
for positive integers 
.
,
,
,
.
we get 
,
,
,
.
,
for positive integers 
,
,
.
we get 
,
,
,
.
achieved when 
,
and 
bisecting 
