<s>[INST] 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.
 [/INST] 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!
</s>