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.
is a rational number.
and other permutations.
, with 
as its incenter and 
as its circumcircle, 
intersects again at 
. Let 
be a point on the arc 
, and 
a point on the segment 
, such that 
. If 
is the midpoint of 
, prove that the meeting point of the lines 
and 
lies on .
such that![\[a+1|b^2+1 \ , \ b+1|a^2+1.\]](http://latex.artofproblemsolving.com/3/c/7/3c7b7da769c4e93f6182c943ed8ed7346297f0a4.png)
+1 w
be reals such that 
Find tha maximum value of 
be reals such that 
and 
Find tha minimum value of 
be a triangle with the incenter , and let the incircle 
of touch 
at points 
respectively. Let 
intersect again at a point 
. Let the lines through and parallel to 
intersect again at points 
respectively. Prove that lies on the common chord of the circumcircle of 
and the circumcircle of 
.
. Prove that:
be a point in the plane, and 
a line not passing through . Evan does not have a straightedge, but instead has a special compass which has the ability to draw a circle through three distinct noncollinear points. (The center of the circle is not marked in this process.) Additionally, Evan can mark the intersections between two objects drawn, and can mark an arbitrary point on a given object or on the plane. over ? to ? be a triangle with 
and with its incircle touching the sides 
and at 
and 
respectively. A point lies on the extension of beyond 
such that 
. Let 
be the midpoint of 
. Prove that the points , , are collinear.
are the reals. Prove that:![\[ \left(\frac {a}{b - c}\right)^2 + \left(\frac {b}{c - a}\right)^2 + \left(\frac {c}{a - b}\right)^2 \ge \frac {1}{2} + \frac {3(ab + bc + ca)}{a^2 + b^2 + c^2}\]](http://latex.artofproblemsolving.com/7/3/4/734032ae618a9addabb3c18c84f8bf2b6e13e588.png)
: are the reals. Prove that: 
such that this ineq is still trues:
, we obtain Dao Hai Long's ineq.
, we obtain Vasc's ineq.
, we obtain my results above is the best? Is this ineq trues with 
such that this ineq is still trues:, we obtain Dao Hai Long's ineq., we obtain Vasc's ineq., we obtain my results above is the best? Is this ineq trues with 
. But 
and are the nicest.
. For 
, the following inequality holds
,
and 
.
there is a very hard inequality
,
equality holds for distinct and nonzero 
. This is an open problem. 
such that this ineq is still trues:
we can prove the inequality by using Cauchy Schwarz Inequality as follows:
and 
Applying the Cauchy Schwarz Inequality, we have![\[ \sum \frac {a^{2}}{(b - c)^{2}}\geq \frac {(a + b + c)^{2}}{\sum (b - c)^{2}} = \frac {x + 2y}{2(x - y)}.\]](http://latex.artofproblemsolving.com/4/1/4/414c37f3d981227908e502bd578e354d348f81dc.png)
so from the above inequality, we get![\[ \frac {2 - k}{3}\sum \frac {a^{2}}{(b - c)^{2}}\geq \frac {(2 - k)(x + 2y)}{6(x - y)}. \eqno (1)\]](http://latex.artofproblemsolving.com/e/2/c/e2ca2d592b56a1e07d8e4249e5ec0671406a6d94.png)
![\[ \sum \frac {a}{b - c}\cdot \frac {b}{c - a} = - 1,\]](http://latex.artofproblemsolving.com/9/9/6/9960e8c583a56b3f1ab69bc4bff807dcfac45dd7.png)
![\[ \sum \frac {a^{2}}{(b - c)^{2}} - 2 = \sum \frac {a^{2}}{(b - c)^{2}} + 2\sum \frac {a}{b - c}\cdot \frac {b}{c - a} = \left( \sum \frac {a}{b - c}\right) ^{2}\geq 0,\]](http://latex.artofproblemsolving.com/b/0/c/b0cc25b8985a37261e87961482950eb2c51de3f0.png)
this inequality gives us that![\[ \frac {k + 1}{3}\sum \frac {a^{2}}{(b - c)^{2}}\geq \frac {2(k + 1)}{3}. \eqno (2)\]](http://latex.artofproblemsolving.com/1/8/2/182b0b53f7f263e2211d5da3025df9a62cb60b0d.png)
![\[ \sum \frac {a^{2}}{(b - c)^{2}} = \frac {2 - k}{3}\sum \frac {a^{2}}{(b - c)^{2}} + \frac {k + 1}{3}\sum \frac {a^{2}}{(b - c)^{2}}\geq \frac {(2 - k)(x + 2y)}{6(x - y)} + \frac {2(k + 1)}{3}.\]](http://latex.artofproblemsolving.com/e/1/e/e1eb65fa1e8b106787a6c3aa27084ed979d055f9.png)
![\[ \frac {(2 - k)(x + 2y)}{6(x - y)} + \frac {2(k + 1)}{3}\geq k + \frac {(4 - 2k)y}{x}.\]](http://latex.artofproblemsolving.com/2/6/6/2666031d8b5dd8ada78c0b8fc1e0f82854bf67ed.png)
![\[ (2 - k)\left[ \frac {x + 2y}{6(x - y)} + \frac {1}{3} - \frac {2y}{x}\right] \geq 0,\]](http://latex.artofproblemsolving.com/c/2/8/c28f103b9c5727402c71383d010619ddccfe4378.png)
![\[ \frac {(2 - k)(x - 2y)^{2}}{2x(x - y)}\geq 0.\]](http://latex.artofproblemsolving.com/0/f/c/0fc3d260e70c49118503b00f2081d23e34089b41.png)
are reals, then
, 
and 
.
and we need to prove that
or
or
for which it's enough to prove that
, which is 
, which is true for . Done!
Prove that
Show that 
.![$$ \frac{ a}{b}+\frac{b}{c}+\frac{c}{a}\geq \sqrt[3]{49}$$](http://latex.artofproblemsolving.com/4/1/7/41747f09ab60b8e1a07de01f23d11985133c0d17.png)
.
we get : 
which is known , but if 
then : 
and it is equivalent to :
of course we may suppose that 
, it is strange because I tried with sophisiticated ways and I failed lol ![\[ \left(\frac{a}{b-c}\right)^2 +\left(\frac{b}{c-a}\right)^2+\left(\frac{c}{a-b}\right)^2+\frac{k(ab+bc+ca)}{(a+b+c)^2} \ge \frac{8+k}{4}\]](http://latex.artofproblemsolving.com/d/9/2/d924963eb3af9bb1f1b0d4d73e5e52a6bc5ab8bd.png)
,![\[ \left(\frac {a}{b - c}\right)^{2} + \left(\frac {b}{c - a}\right)^{2} + \left(\frac {c}{a - b}\right)^{2} \ge \frac {8(ab + bc + ca)}{(a + b + c)^2}\]](http://latex.artofproblemsolving.com/e/9/9/e9928ec3dd1821093e06e0cd02769fcf0ba1484b.png)
![\[ \left(\frac {a}{b - c}\right)^2 + \left(\frac {b}{c - a}\right)^2 + \left(\frac {c}{a - b}\right)^2 + \frac {k(ab + bc + ca)}{(a + b + c)^2} \ge \frac {8 + k}{4}\]](http://latex.artofproblemsolving.com/d/8/1/d81d2b42249e2a6d97b41b4c956369f0afd9198c.png)
