and 
. Prove that
Where 
Where 
vertices is given. To each vertex, a natural number from the set 
is assigned. Prove that there are 
vertices 
such that if the numbers 
are assigned to them respectively, then 
and 
is a parallelogram.
Y by Adventure10, Mango247
Let 
such that 
,prove :
.
plz do not expand(because i want to see a solution different of mine),thx in advance.
such that 
,prove :
.plz do not expand(because i want to see a solution different of mine),thx in advance.
such that 
,
.
![$ \sqrt[3]{abc} \le \frac{a+b+c}{3}=1.$](http://latex.artofproblemsolving.com/f/c/d/fcdee3361c66fbf3696139b3eeaf0d4afeb653ce.png)
![${ \sum\frac{1}{2ab^{2}+1}=\sum (1-\frac{2ab^{2}}{2ab^{2}+1})=\sum (1-\frac{2ab^{2}}{ab^{2}+ab^{2}+1})\geq\sum (1-\frac{2ab^{2}}{3\sqrt[3]{a^{2}b^{4}}})=\sum (1-\frac{2}{3}\sqrt[3]{ab^{2}}})\geq \sum (1-\frac{2}{3}\frac{a+b+b}{3})=\sum (1-\frac{2}{9}(a+2b))=3-\frac{2}{9}(3a+3b+3c)=3-2=1$](http://latex.artofproblemsolving.com/5/2/3/5235814ce188e7049e3cf3ce1e8f06a760bb0c38.png)
and 
are non-negatives such that 
Prove that:![\[ \frac {1}{2 + ab^2} + \frac {1}{2 + bc^2} + \frac {1}{2 + ca^2} \ge 1\]](http://latex.artofproblemsolving.com/7/0/5/70555ed6014bc67521c9ea1e8b3dedeb7e6d0f8b.png)
![$ \sum\frac{1}{2+ab^{2}}=\sum\frac{1}{2}(1-\frac{ab^2}{2+ab^{2}})=\sum\frac{1}{2}(1-\frac{ab^2}{1+1+ab^{2}})\geq\sum\frac{1}{2}(1-\frac{ab^2}{3\sqrt[3]{ab^{2}}})=\sum\frac{1}{2}(1-\frac{1}{3}\sqrt[3]{a^{2}b^4})\geq\sum\frac{1}{2}(1-\frac{1}{3}\frac{ab+ab+b^2}{3})=\frac{1}{2}(3-\frac{1}{9}(a^{2}+b^{2}+c^{2}+2ab+2bc+ac))=\frac{1}{2}(3-\frac{1}{9}(a+b+c)^{2})=1$](http://latex.artofproblemsolving.com/5/2/b/52bd7002205d7c1d3253d7b21d22ed565316e04b.png)
are nonnegative real numbers satisfying 
then
