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.
 A: 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.
A: (1) Consider $$ \Delta =\bigg\{x=(x_i)\in \mathbb{R}^5\bigg| \sum_i\
x_i=1,\ x_i\geq 0\bigg\}
$$
If $\sigma x =\sum_i\ x_ie_{i+1}$ where $e_i$ is a canonical basis
in $\mathbb{R}^5$, then define $$ f: \Delta\rightarrow \mathbb{R},\
f(x)=x\cdot \sigma (x) $$
Define $H=\bigg\{ v\in \mathbb{R}^n\bigg| \sum_i\ x_i=0\bigg\}$. For
$v\in H$, then
$$ \frac{d^2}{dt^2} \bigg|_{t=0}\ f(p+tv) = 2v\cdot \sigma (v) <0 \ \ast $$
(2) Proof of $\ast $ : If $\pi : \mathbb{R}^5\rightarrow H$ is orthogonal
projection, then $\{\pi (e_i)\}$ forms vertex set of tetrahedron $T$
in $H$. And $\sigma$ acts on the set transitively and note that
vertexes has an angle $>\frac{\pi}{2}$.
(3) So $f$ is a strict concave so that it has a global maximum. Here
$$f'(p+tv)\bigg|_{t=0} = p\cdot \{ \sigma (v) +\sigma^{-1}(v)\} =0 $$
 when $p=\frac{1}{5}\sum_{i}\ e_i$.
Hence $f(p)=\frac{1}{5}$
A: 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 $\le\frac15$.
A: Let $y=x+a$, $z=x+a+b$, $w=x+a+b+c$ and $u=x+a+b+c+d$.
Thus, $a$, $b$, $c$ and $d$ are non-negatives and we need to prove that:
$$(x+y+z+u+w)^2\geq5(xw+wz+zy+yu+ux)$$ or
$$(5x+4a+3b+2c+d)^2\geq5(x(x+a+b+c)+(x+a+b+c)(x+a+b)+(x+a+b)(x+a)+(x+a)(x+a+b+c+d)+(x+a+b+c+d)x)$$ or
$$(5x+4a+3b+2c+d)^2\geq5(5x^2+(8a+6b+4c+2d)x+(a+b+c)(a+b)+(a+b)a+a(a+b+c+d))$$ or
$$(4a+3b+2c+d)^2\geq5((a+b+c)(a+b)+(a+b)a+a(a+b+c+d)),$$
which is obvious.
Done!
