AoPS Academy
that has 8097 distinct points with area of a triangle that has 3 points belong to all 
. Prove that there exists a triangle 
that has the area 
contains at least 2025 points that belong to ( each of that 2025 points can be inside the triangle or lie on the edge of triangle )X
is even it's contradiction. I mean 
.
and 
,we can see the only solution is
and it is contradiction.
and 
. So 
.İt means 
.
but 
.İt is contradiction.
we have 2 option. Firstly 
but 
is contradiction. Secondly 
but is contradiction. I mean 
. So 
. If 
but is contradiction. I mean 
. The only solution is 
.
of positive integers, a partition of into two disjoint nonempty subsets 
and 
is 
if the least common multiple of the elements in is equal to the greatest common divisor of the elements in . Determine the minimum value of 
such that there exists a set of positive integers with exactly 
good partitions.
are lengths of sides of some triangle. Prove the inequality
are equal to 
, the original inequality is obvious. Now consider 
. The inequality is equavilent to![\[\sum{\left(\frac{3}{2}(a+b)-\sqrt{a^2+7ab+b^2}\right)}\geq 3\sum{a}-\frac{2(a^2+b^2+c^2)+7(ab+bc+ca)}{a+b+c}\]](http://latex.artofproblemsolving.com/5/3/5/535140526ebcceb34a7182b9a2052a5466033596.png)
![\[\Longleftrightarrow \sum{\frac{5(a-b)^2}{6(a+b)+4\sqrt{a^2+7ab+b^2}}}\geq \frac{\sum{(a-b)^2}}{2(a+b+c)}\]](http://latex.artofproblemsolving.com/d/9/1/d91aa43da4880b2c149444de8de9efe664808a06.png)
, where![\[S_c=\frac{5}{6(a+b)+4\sqrt{a^2+7ab+b^2}}-\frac{1}{2(a+b+c)}\]](http://latex.artofproblemsolving.com/a/0/a/a0a16428e814cc04ae136c832bd9f5fa66b41cae.png)
can be expressed similarly. Assume that 
. First, let's prove that 
, that is![\[\frac{5}{6(c+a)+4\sqrt{c^2+7ca+a^2}}\geq \frac{1}{2(a+b+c)}\]](http://latex.artofproblemsolving.com/b/f/e/bfe152eab6068e62fed9a31f62194df4576afc32.png)
![\[\Longleftrightarrow 2a+2c+5b\geq 2\sqrt{c^2+7ca+a^2}\]](http://latex.artofproblemsolving.com/a/7/a/a7ad69da630fbb898289749f3127f146cbcbdd84.png)
![\[\Longleftrightarrow -20ac+20ab+20bc+25b^2\geq 0\]](http://latex.artofproblemsolving.com/5/7/7/5778081cf324d5de72788a074680542545f4f321.png)
. We can prove 
as the same way. Now it suffices to show that 
, which is![\[\frac{5}{6(a+b)+4\sqrt{a^2+7ab+b^2}}+\frac{5}{6(c+a)+4\sqrt{c^2+7ca+a^2}}\geq \frac{1}{a+b+c}\]](http://latex.artofproblemsolving.com/9/5/5/9550860874cce0c82ea8ed764cfb04704b48ade9.png)
![\[\frac{20}{12a+6b+6c+4\sqrt{a^2+7ab+b^2}+4\sqrt{c^2+7ca+a^2}}\geq \frac{1}{a+b+c}\]](http://latex.artofproblemsolving.com/c/8/1/c81cb96fa425e32ce4d9eae40bbe14a52a00e179.png)
![\[\Longleftrightarrow 4a+7b+7c\geq 2(\sqrt{a^2+7ab+b^2}+\sqrt{c^2+7ca+a^2})\]](http://latex.artofproblemsolving.com/f/2/2/f22ebf4001c3e257356567c9b113d2d8f4cd46d2.png)
![\[(4a+7b+7c)^2\geq 8(2a^2+b^2+c^2+7ab+7ac)\]](http://latex.artofproblemsolving.com/1/3/e/13e4ceee4fc56dc091d99bcfe0fa042c6d6ecc8e.png)
or 
and its permutations .
, 
and 
are non-negatives such that 
. Prove that:![\[\sum_{cyc}\sqrt{a^2+7ab+b^2}\leq\frac{2(a^2+b^2+c^2)+7(ab+ac+bc)}{a+b+c}\]](http://latex.artofproblemsolving.com/2/b/e/2bec411802cbe089dffc05bf1d92ec6936cfce3c.png)
If 
then the inequality is trivial, so we only need to consider 
Now, using the AM-GM inequality, we get
with noting that 
we can rewrite the above inequality as
![$9a\left[\frac{b(a+b+c)}{a+2b}+\frac{c(a+b+c)}{a+2c}- b-c\right] +(a+b+c)\left(\frac{9bc}{b+2c} -2b-c\right) \le $](http://latex.artofproblemsolving.com/0/b/5/0b54c13836da2d9a3def5b710061727ddf5ffa09.png)
![$(b-c)^2\left[\frac{9a^2}{(a+2b)(a+2c)}+\frac{2(a+b+c)}{b+2c} -3\right] \ge 0,$](http://latex.artofproblemsolving.com/4/f/1/4f1f19dc509303a343d6c8c5e3119173abae2cbd.png)
and 
, and are non-negatives such that . Prove that:![\[\sum_{cyc}\sqrt{a^2+b^2}\leq\frac{11(a^2+b^2+c^2)+7(ab+ac+bc)}{3\sqrt2(a+b+c)}\]](http://latex.artofproblemsolving.com/4/3/3/433e1cd5ff26aa0dc20d67a78856870d0e46c439.png)
2)3)![\[\sum_{cyc}\sqrt{4a^2+ab+4b^2}\leq\frac{5(a^2+b^2+c^2)+4(ab+ac+bc)}{a+b+c}\]](http://latex.artofproblemsolving.com/b/7/d/b7dd891c0d169bf766725197b6ca0eaff79ad7a8.png)
(![\[\sum_{cyc}{\sqrt{a^2+7ab+b^2}}\leq\frac{2(a^2+b^2+c^2)+7(ab+bc+ca)}{a+b+c}\]](http://latex.artofproblemsolving.com/3/6/1/36199f2ca50a648b8426a939fddacc25fb365363.png)
and 
