If $a,b,c,d,e,f$ are non negative real numbers such that $a+b+c+d+e+f=1$, then find maximum value of $ab+bc+cd+de+ef$ $(a+b+c+d+e+f)^2=$ sum of square of each number (X)+ $2($ sum of product of two numbers (Y) $)$
$ab+bc+cd+de+ef \le Y$ since all are positive.
Therefore $1\ge X+(ab+bc+cd+de+ef)$
Edit: From AM GM inequality, $$X\ge 6(abcde)^{1/3}$$and for am-gm inequality for a,b,c,d,e,f: $$1\ge 6(abcdef)^{1/6}$$
Hence $X\ge 6.\frac{1}{36}$ and $X\ge \frac{1}{6}$. 
Is this correct?

I will give you an example with 4 variables to show how to solve such problems:
$a+b+c+d=1,b=1-(a+c)-d, ab+bc+cd=b(a+c)+cd=(1-(a+c))(a+c)-d(a+c)+cd=(1-(a+c))(a+c)-ad \le \left(\dfrac{(1-(a+c))+(a+c)}{2}\right)^2-ad \le \dfrac{1}{4}$
when $d=0,a+c=\dfrac{1}{2},b=\dfrac{1}{2}$ you will get maximum value.
