Prove or disprove $ab+bc+cd+da\leq1$ if $a+b+c+d=2$ Non-negative real numbers $a,b,c,d$ are such that $a+b+c+d=2$.  Prove or disprove that
$$ab+bc+cd+da\leq1$$
I see there are multiple equality cases, where $(a,b,c,d)$ is for example $(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2})$, $(\frac{3}{4},\frac{1}{2},\frac{1}{4},\frac{1}{2})$, $(\frac{3}{4},\frac{5}{8},\frac{1}{4},\frac{3}{8})$.
I suspect it's true, and maybe it can be proven with rearrangement, but I have not found a way.
It's reminiscent of Chebyshev's inequality, since the desired inequality is equivalent to
$$4(ab+bc+cd+da)\leq(a+b+c+d)(b+c+d+a)$$
But we cannot assume that $(a,b,c,d)$ and $(b,c,d,a)$ are oppositely ordered.

Let $x = a+c$ and $y=b+d$. Then, we have $xy = (a+c)(b+d) = ab + ad + bc + cd$ which is the quantity we want to bound. So, we want to show that if $x+y = 2$, then $xy \leq 1$.
We just use basic algebra now: $x=2-y$, so $xy = y(2-y)$. Now you can rearrange, $y(2-y) = -(y-1)^2 +1$. Since $(y-1)^2 \geq 0$, it follows that $-(y-1)^2\leq 0$ and so $xy = -(y-1)^2+1 \leq 1$.
You could also just differentiate $-y^2 + 2y$, to see that you have a local maximum when $y=1$.
