[/quote]
,
,
and 
.
is prime, 
with 
and 
be a function such that for every regular 
-
we have 
.
for all reals 
.
be a triangle 
)
are the altitudes of the triangle. The bisector of 
intersects 
at 
.
be a point such that 
and 
.
passes through the midpoint of 
.
& 
.
with 
and at most one of them being zero, show that![\[\frac{1}{\max(a^2,b)}+\frac{1}{\max(b^2,c)}+\frac{1}{\max(c^2,a)}\geq\frac{27}{4}\]](http://latex.artofproblemsolving.com/9/e/1/9e14a9ef2d1a0a98006bcf716fa33c3d9ea29bd5.png)
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
and 
.
Equality holds when 
or 
and 
,
cookies from one of the piles, of which he eats 
and puts 
for which Marica has a winning strategy.
be non-negative real numbers such that 
Prove that
be the circumcenter, symmedian point of 
.
at 
,
at 
,
at 
are concurrent.
Y by Crazy_LittleBoy, Adventure10, Mango247, and 2 other users
Long time ago I have posted the following inequality:
Let
be positive real numbers such that 
Prove that
No solution was posted. But I think it is an interesting inequality and I have found a CS solution, so I revise it here.
Please try on it again.
Let
be positive real numbers such that 
Prove that
No solution was posted. But I think it is an interesting inequality and I have found a CS solution, so I revise it here.
Please try on it again.
and 
.
.
method is the most natural method to solve this inequality.
, 
then:
.
.
,
,
.
.
,
; 
.
.![\[ (a-b)^2(b-c)^2(c-a)^2=q^2-4q^3+2(9q-2)r-27r^2 .\]](http://latex.artofproblemsolving.com/b/4/f/b4f5c57b0df7d34f4730cf6798a075464b29e36c.png)
![\[ a^2b+b^2c+c^2a \le \frac{q-3r+\sqrt{q^2-4q^3+2(9q-2)r-27r^2}}{2} .\]](http://latex.artofproblemsolving.com/1/0/a/10a75d22d485699f7a13c47765272fb20ac8e768.png)
![\[ \frac{1}{r}+\frac{72}{q-3r+\sqrt{q^2-4q^3+2(9q-2)r-27r^2}} \ge 343 .\]](http://latex.artofproblemsolving.com/6/9/3/6933ea8423b7bf636a24c9583d958e1d37b54654.png)
thì bất đẳng thức cần chứng minh tương đương với:![\[ \left(\frac{1029r^2+69r}{343r-1}-q \right)^2-(q^2-4q^3+2(9q-2)r-27r^2) \ge 0. \]](http://latex.artofproblemsolving.com/3/a/4/3a4deadb90eb1759f62347bf063ca1a8e86e847e.png)
![\[ 4q^3-\frac{24r(343r+5)}{343r-1} q +4r+27r^2+\left(\frac{1029r^2+69r}{343r-1}\right)^2 \ge 0. \]](http://latex.artofproblemsolving.com/9/1/e/91eb8dc5b416882d6cfa34c0798e143df50080ee.png)
![\[ 4q^3+8 \sqrt{\left(\frac{2r(343r+5)}{343r-1}\right)^3} \ge \frac{24r(343r+5)}{343r-1} q. \]](http://latex.artofproblemsolving.com/f/a/8/fa85f7fc6f1e46644d49164479f4fbcad80ee34f.png)
![\[ 4+27r+\frac{r(69+1029r)^2}{(343r-1)^2} \ge 8 \sqrt{\frac{8r(343r+5)^3}{(343r-1)^3}}. \]](http://latex.artofproblemsolving.com/d/a/7/da77351f6e10cfff5a5ef3c7fa68fc3a1c328d9a.png)
![\[ 1+\frac{9r((343r+5)^2+108)}{(343r-1)^2} \ge 2\sqrt{\frac{8r(343r+5)^3}{(343r-1)^3}}. \]](http://latex.artofproblemsolving.com/6/0/e/60eb20190ccc11ee457257aae36c5f1bb69305ff.png)
,
và 
![\[ 1+\frac{9(t+5)(4t^2-6t+3)}{343(t-1)} \ge 2\sqrt{\frac{8t^3(t+5)}{343(t-1)}} .\]](http://latex.artofproblemsolving.com/6/5/3/653671e9beb94b7be77b2fe099a4d379b34971bd.png)
![\[ 18t^3+63t^2+50t-104 \ge 7\sqrt{56t^3(t-1)(t+5)}. \]](http://latex.artofproblemsolving.com/2/1/1/211f0cf28eac8df781658397c3d1497e0eb4c14f.png)
![\[ (t-2)^2(324t^4+820t^3-3223t^2+104t+2704) \ge 0 .\]](http://latex.artofproblemsolving.com/f/c/0/fc07f39e1c88788671ade293398798fff44cc8ef.png)
,
và 
và các hoán vị.
