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 
, define 
as the greatest exponent 
such that 
divides 
. Prove that there are infinitely many such that 
and infinitely many 
such that 
.
with circumcircle 
is given. Let 
be the midpoint of 
, 
be the foot of altitude from 
to 
, and 
be the intersection point of 
and 
. Point 
lies on the arc of (the one that does not contain ) such that 
. Prove that 
.
be a trapezium inscribed in a circle 
with diameter . Let be the intersection point of the diagonals and 
. The circle with center 
and radius 
meets at the points 
and 
(where is on the same side of as 
). The line perpendicular to at intersects 
at 
. Prove that 
is perpendicular to 
.![\[ x + y^2 = p^m \]](http://latex.artofproblemsolving.com/e/c/f/ecf553036f22327dd457dc06611b6bc1b0ef072c.png)
![\[ x^2 + y = p^n \]](http://latex.artofproblemsolving.com/5/b/d/5bd672934a395aabd6865ca4b50d8da75c0fe07e.png)
For 
positive integers and 
being a prime.
and 
. Prove that
Equality holds when 
or 
, , and lie on a circle such that 
. The tangent line to at point intersects the tangent line to at and the line 
at and . The circle and the circumcircle of triangle 
intersect again at . Prove that 
. and . Prove that
Equality holds when or 
intersect its circumcircle at 
. The tangets to from 
determine a triangle 
, where 
is relative to and so on. Prove that lines 
are concurrent at a point 
and that belongs to the euler line of . be a cyclic quadrilateral whose opposite sides are not parallel. Suppose points 
lie in the interiors of segments 
respectively, such that 
Let 
intersect 
at 
, and 
intersect 
at 
. Prove that lines 
and 
are either parallel or coincide.
. Prove that:
![$ \sqrt [3]{(a + b)(a + c)(b + c)}\le \frac {(a + b) + (a + c) + (b + c)}{3} = \frac {2.(a + b + c)}{3}\Rightarrow$](http://latex.artofproblemsolving.com/a/b/3/ab3eb1a322f24b326b35dc9263c3389083479d34.png)
then: 
and
![$ (2x + y + z)(x + 2y + z)(x + y + 2z)=[(x+y+z)+x][(x+y+z)+y][(x+y+z)+z]=$](http://latex.artofproblemsolving.com/4/d/1/4d1f445304338426212695c8e7cea4d4bfa1b2ac.png)
![$ =[(x+y+z)^2+(x+y+z)(x+y)+xy][(x+y+z)+z]=$](http://latex.artofproblemsolving.com/9/b/7/9b77896e8c82b76c123cbc963d95ceb81cf0062d.png)
, so 
and we have the inequality 
, 
, 
, this is equivalent to
so we are done.
.
, we have:
, 
and 
and 
.
. 
doesn't need to be the incentre of .
, 
, 
and 
.
.
and 
, therefore 
.
.
, 
and 
. Since is the incentre of , 
, 
and 
.
.
, 
and 
. We want to prove 
.
, and . We want to prove .
and your computation will get more easier.
.
so 
and our inequality is equivalent to:
and 
and multiplying it with 
we get 
q.e.d.
or
( By Bernoulli)
![\[\dfrac14 < \dfrac{(a+b)(b+c)(c+a)}{(a+b+c)^3}\leq\dfrac{8}{27}.\]](http://latex.artofproblemsolving.com/5/b/4/5b4df44828dc4adf330219eba597693f1ce4f6dd.png)
Let , 
, 
for 
(Ravi Substitution). Then the expression in the middle becomes ![\[\dfrac{(2x+y+z)(x+2y+z)(x+y+2z)}{8(x+y+z)^3},\]](http://latex.artofproblemsolving.com/5/2/3/5232afeec06481f04a82f92b3d42bd01180db443.png)
and so the inequality becomes 
Now let 
, 
, 
, so that 
. Then the inequality becomes ![\[2<(1+p)(1+q)(1+r)\leq\dfrac{64}{27},\]](http://latex.artofproblemsolving.com/8/7/7/8775c4356c01090e6c2b79db5d94846b59a057fc.png)
which is much easier to work with.![\[(1+p)(1+q)(1+r) = 2 + (pq+qr+rp+pqr) > 2.\]](http://latex.artofproblemsolving.com/7/4/5/745e8119020d38eec0d2fbeaab28a80422e47815.png)
![\[(1+p)(1+q)(1+r)\leq\left(\dfrac{3+p+q+r}3\right)^3 = \dfrac{64}{27}.\]](http://latex.artofproblemsolving.com/e/b/0/eb04ccc1e163b72d46d84259935d7ae93425bb37.png)
so we need to prove .![$2 = \frac{b+c}{a+b+c} + \frac{c+a}{a+b+c} + \frac{a+b}{a+b+c} \ge 3\sqrt [3]{\frac{(a + b)(a + c)(b + c)}{(a + b + c)^3}} \implies \frac{8}{27} \ge \frac {(a + b)(a + c)(b + c)}{(a + b + c)^3}$](http://latex.artofproblemsolving.com/c/2/b/c2bfbf144e03cb82f3e5ce125f8ebe60ab1aaaad.png)
such that 
. we have to prove 
which is True.
without using Ravi Transformation .
, W.L.O.G.![$4S : = 4[ABC] = \sqrt {(1-2a)(1-2b)(1-2c)} \Longleftrightarrow (1-2a)(1-2b)(1-2c) = 16S^2.$](http://latex.artofproblemsolving.com/b/3/b/b3b420d9fe16d78e895eb5abd8a630be5905be0e.png)
.
The right side is true by AM-GM. For the left side, let 
with 
. Then the inequality becomes 
which is true as 
and 
.
and 
.
Where 
such that :
let 
be the center of its inscribed circle. The internal bisectors of the angles 
meet the opposite sides in 
respectively. Prove that![\[ \frac {1}{4} < \frac {AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac {8}{27}.
\]](http://latex.artofproblemsolving.com/2/d/b/2dbd2b1c292cd1de832fcc26c647a72b6347dc99.png)
.
.
.
Prove that 
Solution:![$$\sum_{cyc}\sqrt{\frac{2}{\frac{1}{x^3}+1}}\leq \sum_{cyc}\sqrt{\frac{2}{2\sqrt{\frac{1}{x^3}}}}=\sum_{cyc}\sqrt[4]{x^3}\leq\sum_{cyc}\frac{3x+1}{4}=3.$$](http://latex.artofproblemsolving.com/4/c/e/4cefa64c1a878b182478812a9dc904534042bcdf.png)
![\[{\frac { \left( a+b \right) \left( b+c \right) \left( c+a \right) }{
\left( a+b+c \right) ^{3}}}\geq \frac{1}{4}+{\frac {5}{36}}\,{\frac {\sqrt {3}r}{
s}}\]](http://latex.artofproblemsolving.com/0/b/e/0bee12e340a326f950516644302fb1ea52e2131f.png)
![\[{\frac { \left( m_a+m_b \right) \left( m_b+m_c \right) \left( m_c+m_a \right) }{
\left( m_a+m_b+m_c \right) ^{3}}}\geq \frac{1}{4}+{\frac {5}{36}}\,{\frac {\sqrt {3}r}{
s}}\]](http://latex.artofproblemsolving.com/e/a/f/eaf96b8292db3e3f393f7e5939f4338ea14633db.png)
where 
and 
be the length of the A-altitude and 
Now, drop an altitude from I to 
This means that 
Now, let line 
be a line parallel to 
Line 
respectively. Now, let 
intersects line 
From similar triangles, ![$$\frac{AI}{AA'}=\frac{AN}{AC}=\frac{AA_1}{AH}=\frac{AH-A_1H}{AH}=\frac{h_a-IM}{h_a}=\frac{h_a-r}{h_a}=\frac{\frac{2[ABC]}{a}-\frac{[ABC]}{s}}{\frac{2[ABC]}{a}}=\frac{\frac{2}{a}-\frac{1}{s}}{\frac{2}{a}}=\frac{\frac{1}{a}-\frac{1}{a+b+c}}{\frac{1}{a}}=\frac{b+c}{a+b+c}.$$](http://latex.artofproblemsolving.com/6/9/8/698ff27ac9b0065c98024799c0b962d91b27ca2a.png)
From the Lemma, it now suffices to prove that 
Let 
and 
Now, from the AM-GM, 
Now, let 
and 
We can easily find that 
This means that 
Combining, we obtain 
as desired.
and 
Prove that 
