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 $ab+bc+cd+de+ef$
Solution 1:
$$a+b+c+d+e+f=1\implies (a+c+e)(b+d+f)\le \frac14 \\ \implies (ab+bc+cd+de+ef)+(ad+af+cf+be)\le \frac14$$ We can show the first bracket can attain full value by say $a=b=\frac12$.
Solution 2:
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.