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.
be positive reals. Prove that![\[ \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{2b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{2c^2 + (a+b)^2} \geq 0. \]](http://latex.artofproblemsolving.com/6/9/3/6937791380a8642edca3e773dc72dbf3f0e0f206.png)
be a perfect square then, is it true that it must be of the form
be points on the sides 
respectively of a triangle 
. Suppose that 
and 
meet at 
and meet at 
, and 
and meet at 
, such that 
, and 
. Compute the ratio of the area of 
to the area of 
.
and 
intersect at different points 
and 
. The straight lines tangents to that pass through and intersect at 
. Let 
be a point on that is out of . The 
line intersects at 
again, the 
line intersects again to in 
and the line 
intersects again to the circumference in 
. Prove that the 
line passes through the midpoint of the 
segment.
are positive real numbers such that 
, prove that ![\[ (1+a)(1+b)(1+c) \geq 8 (1-a)(1-b)(1-c) . \]](http://latex.artofproblemsolving.com/0/b/e/0bedd9ed8362ac5202845f8a4d0d24519423e9b4.png)
with 
, points 
are such that 
and 
and 
, and 
and 
, as shown on the picture. Suppose that 
is a right angle. Prove that the points 
are collinear.
. Prove that![\[
\sum_{\text{cyc}} \frac{x^3}{y^2 + z^2} \geq \frac{x + y + z}{2}
\]](http://latex.artofproblemsolving.com/9/0/3/903220c471bdd3028dda2fcb13874a4bcf752872.png)
without using the Rearrangement Inequality or Chebyshev's Inequality. be an acute angled triangle.And altitudes 
and 
intersects at point 
.Let 
be a point on ray such that 
.Circumcircle of 
intersects line 
at and so prove that 
,![\[\frac{1}{a+\frac{1}{b}+1}+\frac{1}{b+\frac{1}{c}+1}+\frac{1}{c+\frac{1}{a}+1}\ge \frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}. \]](http://latex.artofproblemsolving.com/4/3/1/431b0ed375e037ee3d19e48a2bb4099f12f87784.png)
Proposed by David Stoner
such that![\[ x f(x + xy) = x f(x) + f \left( x^2 \right) f(y) \quad \forall x,y \in \mathbb{R}.\]](http://latex.artofproblemsolving.com/b/0/f/b0f1a1d3b9de5d23f1b5a2e9be3b88d8bd575883.png)
is not divisible by 
for any positive integer 
.
, s.t :
formed by using the digits 
once each such that the number 
is divisible by 
for 
.
and be given circles. Let 
on and 
on be fixed points. If chord 
of is parallel to chord 
of , find the locus of the midpoint of 
.
be the centers of 
Let 
be the midpoints of 
and 
the midpoints of 
i.e. the projections of on 
respectively. Let be the midpoint of 
and the perpendicular bisector 
of cuts 
at 
respectively. Since 
is midparallel of 
then is midpoint of 
and 
which means that 
and 
are equidistant from goes through the midpoint 
of 
locus of is the circle with center and radius 
,such that 
P
