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.

We have: $ab+bc+cd+da = (a+c)(b+d) \le \dfrac{((a+c)+(b+d))^2}{4}= \dfrac{(a+b+c+d)^2}{4}= 1$.
