Just $m$

Let (x+1) = $p^{2}$ & (3-x) = $q^2$ .

We have $p^{2} + q^{2} = 4$ .

Now by AM of 2nd powers greater than 2nd power of AM's & AM-GM we have,

$\frac{p^{2} + q^{2}}{2} \ge (\frac{p+q}{2})^{2} \ge pq$

So we get two relations namely,

$(p+q)\ge 2\sqrt2)$ ...................... (1)

$2pq\ge 4$ .....................................(2) [Since $p^{2} + q^{2} = 4$ ]

So our expression = $p+q+2pq\le m+1$

or, $m\ge (p+q) + (2pq) - 1\ge 2\sqrt2 + 4 - 1= 2\sqrt2 + 3 = 5.82$ [From (1) & (2)]