Maximum value of function

[man111]

Given [tex]a,b,c,d,e,f\geq 0[/tex] and [tex]a+b+c+d+e+f=1,[/tex] Then maximum value of [tex]ab+bc+cd+de+ef[/tex]

I have tried Using Multiplier method::

Let [tex]f(a,b,c,d,e,f) = ab+bc+cd+de+ef-\lambda(a+b+c+d+e+f-1)[/tex]

Now for [tex]\max[/tex]

Put [tex]\displaystyle \frac{df}{da} = 0\Rightarrow b-\lambda = 0\Rightarrow b=\lambda[/tex]

Put [tex]\displaystyle \frac{df}{db} = 0\Rightarrow a+c-\lambda = 0\Rightarrow a+c=\lambda[/tex]

Put [tex]\displaystyle \frac{df}{dc} = 0\Rightarrow b+d-\lambda = 0\Rightarrow b+d=\lambda[/tex]

Put [tex]\displaystyle \frac{df}{dd} = 0\Rightarrow c+e-\lambda = 0\Rightarrow c+e=\lambda[/tex]

Put [tex]\displaystyle \frac{df}{de} = 0\Rightarrow d+f-\lambda = 0\Rightarrow d+f=\lambda[/tex]

Put [tex]\displaystyle \frac{df}{df} = 0\Rightarrow e-\lambda = 0\Rightarrow e=\lambda[/tex]

So we get [tex]\displaystyle a=\lambda\;\;, b=\lambda, c=0\;,d=0,e=\lambda, f=\lambda[/tex]

Put all values in [tex]a+b+c+d+e+f=1[/tex]

So we get [tex]\displaystyle \lambda = \frac{1}{4}[/tex]

So we get [tex]\max(ab+bc+cd+de+ef) = \lambda^2+\lambda^2=2\lambda^2 = \frac{2}{16} = \frac{1}{8}[/tex]

But answer given is [tex]\displaystyle \frac{1}{4},[/tex] But i did not understand where I am doing wrong.

Thanks

[nevrodermit]

[tex]ab+bc+cd+de+ef \le (a+c+e)(b+d+f) \Leftrightarrow ab+bc+cd+de+ef \le ab+ad+af+cb+cd+cf+eb+ed+ef \Leftrightarrow 0\le ad+af+cf+eb[/tex] which is true since the numbers are non-negative. Equality occurs when [tex]ad+af+cf+eb=0[/tex].
From AM-GM:
[tex](a+c+e)(b+d+f)\le (\frac{a+c+e+b+d+f}{2})^2=\frac{1}{4}[/tex]. So the maximum value is [tex]\frac{1}{4}[/tex].
Equality occurs when [tex]a+c+e=b+d+f=\frac{1}{2}[/tex] and [tex]ad+af+cf+eb=0[/tex]. From the fact that the numbers are non-negative and that [tex]ad+af+cf+eb=0[/tex] we must have [tex]ad=af=cf=eb=0[/tex].
Case 1: [tex]a=0[/tex] then [tex]cf=eb=0[/tex]. At least one of [tex]c,e[/tex] is not zero, otherwise [tex]a+c+e=0[/tex].
1) Let [tex]c=0, e =\frac{1}{2}[/tex]. Also [tex]b=0[/tex], [tex]d+f=\frac{1}{2}[/tex].
The other cases are similar.