any solution???

We claim that the answer is $2 - \sqrt3 > \frac14.$ First let us show that this is attainable. Let $\triangle ABC$ be an equilateral triangle of side length $1$. Then, let $D$ be the unique point such that $BD = 1, DA = DC, $ and $ABCD$ is a convex quadrilateral. In this case, $abcd = 1 \cdot 1 \cdot 2 \cos 15 \cdot 2 \cos 15 = 2 - \sqrt 3$. Let us now show that this is optimal. We will redefine "convex" to permit angles of the quadrilateral to be $180$ degrees. Call a convex quadrilateral satisfying the conditions of the problem a tapir if it has the maximum possible area. We wish to show that all tapirs have area $ \le 2 - \sqrt 3,$ and we already know that all tapirs have area $\ge 2 - \sqrt 3$. We will call a pair of points $(P_1, P_2)$ tasty if the distance between them is one. Lemma 1. No tasty quadrilateral has three collinear vertices. Proof. Suppose that $A, B, C$ were collinear. Then, we have that $AD \cdot DC \cdot CB \cdot BA \le 1 \cdot 1 \cdot \frac14 (AB + BC)^2 \le 1 \cdot 1 \cdot \frac14 \cdot 1 = \frac14 < 2 - \sqrt3,$ contradiction to the fact that $ABCD$ was a tapir. $\blacksquare$ Lemma 2. For every tapir $ABCD,$ we have that $(A, C)$ and $(B, D)$ are both tasty. Proof. To see this, start with an arbitrary tapir $ABCD$. Suppose that $(A, C)$ was not tasty. If $(A, D)$ is also not tasty, then we can consider the effect of rotating $A$ away from $D$ about $B$. In this way, $\angle ABD, \angle ABC$ are both increasing. Observe that the lengths of $AB, BC, CD$ are preserved by the process, while the length of $AD$ is increased (by the Law of Cosines). Hence, since $ABCD$ was a tapir, this process must somehow break some condition of the problem. Since we assumed $(A, C), (A, D)$ were both not tasty, the only way a condition can break is if $A, B, C$ are collinear. However, this is a direct contradiction to Lemma $1.$ Therefore, we must have that $(A, D)$ is tasty. By similar reasoning, we can establish that $(A, B), (C, B), (C, D)$ are all tasty. However, this implies that $ABCD$ is a rhombus of side length $1$, and so one of $AC, BD$ must be greater than $1$, contradiction to the fact that $ABCD$ is a tapir. $\blacksquare$ Lemma 3. All tapirs have at least one side of length $1$. Proof. Assume the contrary, for contradiction. Let $\theta_1, \theta_2$ denote $\angle BDA, \angle CDB$ respectively. By Lemma $1$, we know that $\theta_1, \theta_2 >0.$ By Lemma $2$, we know that $BD = 1.$ Consider the effect of rotating $B$ about $C$ in a way that decreases $\theta_2.$ This process clearly preserves $c, d.$ Consider $a^2 b^2 = (d^2 + 1 - 2d \cos \theta_1) (c^2 + 1 - 2c \cos \theta_2)$ as a function of $\theta_1.$ The derivative of this function is $-(d^2 + 1 - 2d \cos \theta_1) \cdot 2c \sin \theta_2+(c^2+1-2c \cos \theta_2) \cdot 2d \sin \theta_1$ by the Product Rule. Since $ABCD$ was a tapir, this derivative must be zero, as otherwise rotating $DB$ about $D$ in some direction increases $abcd.$ Setting this equal to zero, and recalling that $d^2 + 1 - 2d \cos \theta_1 = a^2, c^2 + 1 - 2c \cos \theta_2 = b^2,$ we know: $$2a^2 c \sin \theta_2 = 2b^2 d \sin \theta_1,$$which yields $$\frac{c}{d} \cdot \frac{\sin \theta_2}{\sin \theta_1} = \frac{b^2}{a^2}.$$By Sine Law in $\triangle CDA$, this implies that $E =BD \cap AC$ satisfies $\frac{CE}{EA} = \frac{b^2}{a^2}$ and so therefore $BE$ is the $B-$symmedian of $\triangle ABC.$ Analogously, we can obtain that $CE$ is the $C-$symmedian of $\triangle BCD$. The two previous results then yield that $ABCD$ is a cyclic harmonic quadrilateral. Since $AC = BD = 1$, cyclicity yields that $ABCD$ is actually an isosceles trapezoid. Let $EA = EB = x, EC = ED = 1-x$ WLOG, and set $\angle BEC = \theta.$ Then, $abcd$ can be easily computed as: $$4 \cos^2 (\frac{\theta}{2}) x (1-x) \cdot (x^2 + (1-x)^2 - 2x(1-x) \cos \theta).$$Noting that $4 \cos^2 (\frac{\theta}{2}) = 2 \cos \theta + 2,$ we can rewrite the above as: $$[(2 \cos \theta+2) x(1-x)] \cdot [1 - (2 \cos \theta + 2) x (1-x)].$$Letting $t = (2 \cos \theta + 2)x(1-x)$, the above is just $t(1-t) \le \frac14 < 2 - \sqrt 3.$ This contradicts the fact that $ABCD$ is a tapir. $\blacksquare$ By Lemmas $1, 2, 3,$ it suffices to show that all tapirs satisfying $CA = AB = BD = 1$ have $abcd \le 2 - \sqrt 3.$ Let $P$ be the point such that $\triangle APB$ is equilateral, and $P, C, D$ are all on the same side of $AB.$ The conditions of the problem imply that $\angle DBA, \angle CAB \le 60.$ Note that this implies that $CD \le 1$ for free. Case 1. $P \in \{C, D\}$ Suppose that $P = C$ by symmetry. Let $\angle DBA = 2 \theta$ for some $0 \le \theta \le 30.$ Then, $abcd = 2 \sin \theta \cdot 2 \sin (30 - \theta) = 2(\cos (2 \theta - 30) - \cos 30).$ From here, it is clear that $abcd$ is maximized at $\theta = 15$, which yields $abcd = 2 - \sqrt 3 \le 2 - \sqrt3$. Hence, the case is resolved. Case 2. $P \notin \{C, D\}$ Let $\angle CAB =2 \alpha, \angle DBA = 2 \beta,$ where $0 \le \alpha, \beta \le 30.$ Then, we can easily find $AD, BC$ to be $2 \sin \beta, 2 \sin \alpha$ respectively. By the Pythagorean Theorem, it's easy to find that $c = \sqrt{(\cos 2 \alpha + \cos 2 \beta - 1)^2 + (\sin 2 \beta - \sin 2 \alpha)^2}$. Now, consider $bcd$ as a function of $\alpha$. By the same rotating argument used before, its derivative must be zero, and so we have: $$2 \cos \alpha \cdot c + 2 \sin \alpha \cdot \frac{\partial c}{\partial \alpha} = 0. \qquad (1)$$Observe that: $$\frac{\partial c}{\partial \alpha} = \frac{2(\cos 2 \alpha + \cos 2 \beta - 1) (-2 \sin 2 \alpha) + 2(\sin 2 \beta - \sin 2 \alpha)(-2 \cos 2 \alpha)}{2c},$$and so by $(1)$ we get: $$4 \cos \alpha \cdot c^2 + 2(\cos 2 \alpha + \cos 2 \beta - 1) (-2 \sin 2 \alpha) + 2(\sin 2 \beta - \sin 2 \alpha)(-2 \cos 2 \alpha) = 0. \qquad (2)$$Analogously, we obtain that: $$4 \cos \beta \cdot c^2 + 2(2 \cos 2 \alpha + \cos 2 \beta -1)(-2 \sin 2 \beta)+2(\sin 2 \beta - \sin 2 \alpha)(-2 \cos 2 \beta) = 0. \qquad (3)$$Now, suppose that $\alpha > \beta.$ Then, as $\cos \alpha < \cos \beta, \cos 2 \alpha + \cos 2 \beta \ge 1, \sin 2 \alpha > \sin 2 \beta, \cos 2 \alpha < \cos 2 \beta,$ it's clear that the LHS of $(2)$ is strictly less than that of $(3)$. However, this is clearly a contradiction since the RHS's of $(2), (3)$ are equal. Therefore, $\alpha \le \beta,$ and analogously we get $\beta \le \alpha.$ In conclusion, we obtain that $\alpha = \beta.$ Now, observe that $abcd = 2 \sin \alpha \cdot 2 \sin \alpha \cdot (2 \cos 2 \alpha - 1),$ which is equal to: $$4 \sin^2 \alpha \cdot (1 - 4 \sin^2 \alpha).$$Hence, if we let $\gamma = 4 \sin^2 \alpha,$ we get $abcd = \gamma (1 - \gamma) \le \frac14 < 2 - \sqrt3.$ However, this contradicts the fact that $ABCD$ is a tapir. Hence, Case $2$ is actually absurd. As we've exhausted all cases, we conclude that all tapirs have $abcd$ equal to $2 - \sqrt 3.$ In fact, it's easy to see that we have actually proven that all tapirs are the same up to rotation and relabeling of vertices. $\square$