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?

$$a+b+c+d+e+f=1\implies (a+c+e)(b+d+f)\le \frac14 \\ \implies (ab+bc+cd+de+ef)+(ad+af+cf+be)\le \frac14$$
We can show the first bracket can attain full value by say $a=b=\frac12$. 
