This was fun.
We will in fact prove a stronger statement, that the sum of the products of opposite sides of such a quadrilateral is no more than $2\sqrt{5}$. The result then follows from an application of $AM-GM$. We will work with the algebraic analogue of this problem:
Let $a, b, c, d, x$ be positive reals with $a+b=c+d=2$ and $0\le x\le 1$. Prove that: $\sqrt{(a^2+b^2+2abx)(c^2+d^2+2cdx)}+\sqrt{(a^2+d^2-2adx)(b^2+c^2-2bcx)}\le 2\sqrt{5}$
(Note: here, $x$ is taken to be the nonnegative cosine of the angle between the two diagonal, WLOG supposing that the angle between the segments with length $a$ and $b$ is acute.)
First, we manipulate the expressions under the radicals, call them $L$ and $R$: $\begin{align*} L&=(a^2+b^2+2abx)(c^2+d^2+2cdx) \\ &=(a^2+b^2)(c^2+d^2)+2x(ab(c^2+d^2)+cd(a^2+b^2))+4x^2abcd \\ &=(ac-bd)^2+(ad+bc)^2+2x(ad+bc)(ac+bd)+x^2[(ac+bd)^2-(ac-bd)^2] \\ &=[x(ac+bd)+(ad+bc)]^2+(1-x^2)(ac-bd)^2 \end{align*} $ In a similar manner, we see that: $ R=[x(ac+bd)-(ab+cd)]^2+(1-x^2)(ac-bd)^2$ Now let $p=x(ac+bd)+\frac{(ad+bc)-(ab+cd)}{2}$ and $q=|ac-bd|\sqrt{1-x^2}$. Using the fact that $\frac{ad+bc+ab+cd}{2}=\frac{(a+c)(b+d)}{2}=2$, our inequality can be rewritten as: $\sqrt{(p+2)^2+q^2}+\sqrt{(p-2)^2+q^2}\le 2\sqrt{5}$ We will now prove the key lemma:
Lemma: $p^2+5q^2\le 5$
Proof: Using the fact that $ad+bc-(ab+cd)=(a-c)(d-b)=4(a-1)(b-1)$, this inequality can be rewritten as: $5[a(2-a)-b(2-b)]^2(1-x^2)+[x(a(2-a)+b(2-b))+2(a-1)(b-1)]^2\le 5$ Upon the translation $a\Rightarrow 1+y, b\Rightarrow 1+z$ for $y, z\in [-1, 1]$, this simplifies to: y, z\ge 9$5(y^2-z^2)^2(1-x^2)+[x(2-y^2-z^2)^2-2yz]^2\le 5$ $\Leftrightarrow 4x^2(1+3y^2z^2-y^2-y^4-z^2-z^4)+4xyz(2-y^2-z^2)+5y^4+5z^4-6y^2z^2\le 5$
Let this expression be $F(x, y, z)$. From this formulation, we see that since $F(x, y, z)\le F(x, |y|, |z|)$, we can say $y, z\ge 0$ WLOG. Now, we take two cases:
Case 1: $x\le \frac{3}{4}$. In this case, note that: $\begin{align*}F(x, 1, yz)-F(x, y, z)&= (5-4x^2)(1-y^4)(1-z^4)-(4x^2+4xyz)(1-y^2)(1-z^2) \\ &=(1-y^2)(1-z^2)[(5-4x^2)(1+y^2)(1+z^2)-4x^2-4xyz] \\ &\ge (1-y^2)(1-z^2)[(5-\frac{9}{4})(1+y^2)(1+z^2)-\frac{9}{4}-3yz] \\ &= (1-y^2)(1-z^2)[\frac{1}{2}+\frac{11}{4}(y-z)^2+\frac{5}{2}yz+\frac{11}{4}y^2z^2] \\ &\ge 0 \end{align*}$
It follows that it suffices to prove the inequality for $z=1$. But we see that: $\begin{align*}F(x, y, 1)&=4x^2(-y^4+2y^2-1)+4xy(1-y^2)+5y^4-6y^2+5 \\ &=5-5y^2(1-y^2)-[2x(1-y^2)-y]^2 \\ &\le 5 \end{align*} $ And so this case is complete.
Case 2: $x>\frac{3}{4}$. In this case, note that: $4xyz(2-y^2-z^2)\le 4xyz(2-2yz)=8xyz(1-yz)\le (4x^2+4)yz(1-yz)$ Therefore, it suffices to show that: $4x^2(yz(1-yz)+3y^2z^2+1-y^4-z^4-y^2-z^2)+4yz(1-yz)+5y^4-6y^2z^2+5z^4\le 5$ This last function is linear in $x^2$, so its maximum occurs at the endpoints of its domain. Therefore, it suffices to check the inequality for $x=\frac{3}{4}$ and $x=1$. The latter gives: $y^4+z^4-2y^2z^2-y^2-z^2+2yz+4\le 5$ $\Leftrightarrow (y^2-z^2)^2-(y-z)^2\le 1$ which is true because both terms are in $[0, 1]$. Finally, for $x=\frac{3}{4}$ the inequality becomes: $\frac{11}{4}y^4+\frac{11}{4}z^4-\frac{11}{2}y^2z^2-\frac{9}{4}(y^2+z^2-yz)+\frac{9}{4}\le 5$ $\Leftrightarrow \frac{11}{4}(y^2-z^2)^2-\frac{9}{4}(y^2+z^2-yz)\le \frac{11}{4}$ Which is true because $y^2+z^2-yz\ge 0$.
Now, back to the main inequality. The lemma can be rewritten as: $(p^2+q^2+4)^2-16p^2\le (6-p^2-q^2)^2$ $\Leftrightarrow [(p+2)^2+q^2][(p-2)^2+q^2]\le (6-p^2-q^2)^2$ Because $p^2+q^2<5<6$, this implies that: $\sqrt{[(p+2)^2+q^2][(p-2)^2+q^2]}\le 6-p^2-q^2$ $\Leftrightarrow [\sqrt{(p+2)^2+q^2}+\sqrt{(p-2)^2+q^2}]^2\le 20$ Finally, taking the square root of this gives the desired result. $\Box$
Equality holds when $p=0, q=1$, which corresponds to $(a, b, c, d)=(2, 1, 0, 1)$ and analogous tuples.