Finding the maxima of a multivariable function using Lagrange's Multipliers I'm practicing Lagrange Multipliers (LM)$^{[1]}$ with the following self-made question:

Given $a + b + c + d + e = 1$, where $a, b, c, d, e \notin R^-$. Find the maximum value of $ab + bc + cd + de$

I already know that the answer is $1/4^{[2]}$. But, as an exercise, I want to use LM.

My Attempt:
Let, $$ f(a, \ ... \ ,e) = ab + bc + cd + de \\
g(a, \ ... \ ,e) = a + b + c + d + e = 1 $$
Then, define: $$ \mathcal{L}(a, \ ... \ ,e, \lambda) = f(a, \ ... \ ,e) - \lambda \cdot [g(a, \ ...\ ,e) - 1] \\
\therefore \mathcal{L}(a, ... ,e, \lambda) = ab + bc + cd + de - \lambda \cdot [a + b + c + d + e - 1] $$
Now, $\nabla\mathcal{L} = 0$ would give maxima / minima. On partial differentiation, we get,
$$ \begin{align}
b = \lambda \qquad (from \ \ \frac{\delta\mathcal{L}}{\delta a}) \tag 1\\ 
d = \lambda \qquad (from \ \ \frac{\delta\mathcal{L}}{\delta e}) \tag 2\\ 
a + c = \lambda \qquad (from \ \ \frac{\delta\mathcal{L}}{\delta b}) \\ 
c + e = \lambda \qquad (from \ \ \frac{\delta\mathcal{L}}{\delta d}) \\ 
b + d = \lambda \qquad (from \ \ \frac{\delta\mathcal{L}}{\delta c}) \tag 3
\end{align}$$
Here, equations $(1)$, $(2)$ and $(3)$ seems contradicting. Why it is so?


Note: I tested the same approach with $2$, $3$ and $4$ variables and it gave me correct results. Why so?

References:
[1]: Lagrange multipliers, examples - Khan Academy
[2]: 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: $\color{brown}{\textbf{Task transformations.}}$
Let
$$p=a+c,\quad q=e+c,\tag1$$
then the task is to maximize the function
$$f(b,p,d,q\,|\,c) = bp+dq\tag2$$
under the conditions
\begin{cases}
b+p+d+q = 1+c\\
c\in[0,1],\quad b\in[0,1-c],\quad d\in[0,1-c]\\
p\in[c,1],\quad q\in[c,1],\tag3
\end{cases}
Denote
$$g(c) = \max\limits_{b,p,d,q} f(b,p,d,q\,|\,c),\tag4$$
then
$$\max\limits_{a,b,c,d,e} ab+bc+cd+de = \max g(c).\tag5$$
$\color{brown}{\textbf{Parametric maximum among the trivial cases.}}$
Taking in account the symmetry of the task (the pair $(b,p)$ is symmetric to the pair $(d,q)$), it suffices to consider the next cases.
$\mathbf{1.\quad f(0,p,d,q\,|\,c) = dq.}$
The nesessary conditions of the parametric maximum are
$$p=c,\quad d+q=1,\quad d\in[0,1-c],\quad q\in[c,1],$$
then
$$\max f(0,p,d,q\,|\,c) = \max f(0,c,d,q\,|\,c) = \max\limits_{q\in[c,1]} q(1-q) = c(1-c).\tag6$$
$\mathbf{2.\quad f(b,c,d,q\,|\,c) = bc+dq.}$
The nesessary conditions of the parametric maximum are
$$b+d+q=1,\quad b\in[0,1-c],\quad d\in[0,1-c],\quad q\in[c,1].$$
Since
$$\max f(1-c,c,0,c\,|\,c) = c(1-c),$$
$$\max f(0,c,d,q\,|\,c) = \max\limits_{q\in[c,1]} q(1-q) = c(1-c),$$
$$\max f(b,c,1-b-q,q\,|\,c) = bc+\max\limits_{q\in[c,1-b]} q(1-b-q) = bc+c(1-b-c)=c(1-c),$$
then the least value of $(2)$ among the trivial cases is
$$g_v(c) = c(1-c).\tag7$$
$\color{brown}{\textbf{Parametric conditional maximum.}}$
In accordance with the LM method, the stationary points of the function $(2)$ under the condition $(3.1)$ can be defined via the function
$$F(\lambda,b,p,d,q\,|\,c) = bp+dq + \lambda(b+p+d+q-1-c)$$
from the system of the conditions
$F'_\lambda = F'_b = F'_p = F'_d = F'_q = 0,$ or
\begin{cases}
b+p+d+q=1+c\\
p+\lambda = b+\lambda = d+\lambda = q+\lambda = 0,
\end{cases}
with the solution
$$b=p=d=q = \dfrac{1+c}4,\quad\text{if}\quad c\in\left[0,\dfrac13\right],$$
$$g_s(c) = f\left(\dfrac{1+c}4,\dfrac{1+c}4,\dfrac{1+c}4,\dfrac{1+c}4\,\Big|\ c\right) 
= \dfrac{(1+c)^2}8,\quad c\in\left[0,\dfrac13\right].\tag8$$
$\color{brown}{\textbf{The global maximum.}}$
Taking in account $(7),(3),(8),(5),$ one can get
$$G_v = \max\limits_{c\in[0,1]} g_v(c) = \dfrac14,$$
$$G_s = \max\limits_{c\in[0,\,^1/_3]} g_s(c) = \dfrac{(1+\,^1/_3)^2}8 = \dfrac29,$$
$$\color{brown}{\mathbf{\max\limits_{a+b+c+d+e = 1} ab+bc+cd+de = \dfrac14.}}$$
A: The KKT conditions cannot be solved because $x^*$ is not the optimal point using the constraints you have implemented in your Lagrangian. If we do not implement the inequality constraints
$$
\begin{align}
a \geq 0 \\
b \geq 0 \\ 
c \geq 0 \\
d \geq 0 \\
e \geq 0 
\end{align}
$$
then we can let 
$$a = t, b = t, e = -t, d = -t \text{ and } c=1$$
such for all $t$ the constraint $g(a, \dots, e) = 0$ is satisfied and our function
$$f(a, \dots, e) = t^2 +t - t + t^2 = 2t^2$$ 
is unbounded. We can just let $t\rightarrow \infty$ and our function does not permit a solution. 
As @Toni says we must implement also the inequality constraints $a, b, c, d, e \geq 0$ in the Lagrangian. This yields
$$
\begin{align}
\mathcal{L}(a, b, c, d, e, \lambda, \mu_a, \mu_b, \mu_c, \mu_d, \mu_e) &= 
ab + bc + cd + de \\ 
&+ \lambda(a+b+c+d+e-1) \\
& + \mu_a a + \mu_b b + \mu_c c + \mu_d d + \mu_e e
\end{align}
$$
As you say optimality implies $\nabla \mathcal{L} = 0$
$$
\begin{align}
a + \lambda + \mu_a = 0\\
a + c + \lambda + \mu_b = 0 \\
b + d + \lambda + \mu_c = 0 \\
c + e + \lambda + \mu_d = 0 \\
e + \lambda + \mu_e = 0
\end{align}
$$
This more flexible system can be solved for $a=b= \frac{1}{2}, c=d=e=0$.
$$
\begin{align}
a + \lambda = 0\\
a  + \lambda = 0 \\
b + \lambda + \mu_c = 0 \\
\lambda + \mu_d = 0 \\
\lambda + \mu_e = 0
\end{align}
$$
We find no contradiction. Note that $\mu_a = \mu_b = 0$ as these inequailities are inactive for $a, b > 0$.  
