If $0⩽x⩽y⩽z⩽w⩽u$ and $x+y+z+w+u=1$, prove $xw+wz+zy+yu+ux⩽\frac15$ If $0⩽x⩽y⩽z⩽w⩽u$ and $x+y+z+w+u=1$, prove$$
xw+wz+zy+yu+ux⩽\frac15.
$$


I have tried using AM-GM, rearrangement, and Cauchy-Schwarz inequalities, but I always end up with squared terms. For example, applying AM-GM to each pair directly gives$$
x^2+y^2+z^2+w^2+u^2 ⩾ xw + wz + zy + yu + ux,
$$
but I cannot seem to continue from here or use $x + y + z + w + u = 1$. Other inequalities like Chebyshev's rely on the multiplied pairs to be in order from least to greatest or vice versa, so I am stuck here.
It is quite easy to solve this problem with the Lagrange multipliers

Calling 

$$
L(x,y,z,w,u,\lambda) = \frac 15 -(xw+wz+zy+yu+ux)+\lambda(x+y+z+w+u-1)
$$

The stationary points are the solutions for

$$
\nabla L = (L_x,L_y,L_z,L_w,L_u,L_{\lambda})=0
$$

or

$$
\lambda -u-w = 0\\
\lambda -u-z = 0\\
\lambda -w-y = 0\\
\lambda -x-z = 0\\
\lambda -x-y = 0\\
u+w+x+y+z-1 = 0
$$

Solving this linear system we obtain

$$
x = y = z = w = u = \frac 15,\;\lambda = \frac25
$$

This point is the only tangent point between the surface $g(x,y,z,w,u)=\frac 15 -(xw+wz+zy+yu+ux)$ and the hyperplane $\Pi =x+y+z+w+u-1 = 0$

and  $g(x,y,z,w,u) \ge 0$ is located into one of the  semi-spaces delimited by $\Pi$ and also at the tangency point we have $g = 0$ with the values found before.

NOTE

This formulation is a short hand for

$$
L(x,y,z,w,u,\lambda,\epsilon) = \frac 15 -(xw+wz+zy+yu+ux)+\lambda_1(x+y+z+w+u-1)+\lambda_2(x-\epsilon_1^2)+\lambda_2(y-x-\epsilon_2^2)+\lambda_4(z-y-\epsilon_3^2)+\lambda_5(w-z-\epsilon_4^2)+\lambda_6(u-w-\epsilon_5^2)
$$

and the result should be the same as you can verify with a little patience.
The case where $xw+wz+zy+yu+ux$ is maximal will be where $x=y=z=w=u$. Here, they must all be $\frac15$, and plugging into the equation gives $5(\frac15)^2=\frac15$. So the expression is $\le\frac15$.