Method 1: Residues
Consider the contour integral
$$\oint_C dz \frac{e^{i k z}}{(1+z^2)^2}$$
where $C$ is a semicircle in the upper half plane; here, $k>0$. Then by the residue theorem and Jordan's lemma:
$$\begin{align}\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^2} &= i 2 \pi \operatorname*{Res}_{z=i} \frac{e^{i k z}}{(1+z^2)^2}\\ &= i 2 \pi \left [\frac{d}{dz} \frac{e^{i k z}}{(z+i)^2} \right ]_{z=i}\\ &= i 2 \pi \left [\frac{i k\, e^{i k z}}{(z+i)^2} - \frac{2 e^{i k z}}{(z+i)^3} \right ]_{z=i}\\ &= \frac{\pi}{2} (k+1) e^{-k}\end{align}$$
For $k \lt 0$, $C$ is a semicircle in the lower half plane; for the same reasons as above, we have:
$$\begin{align}\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^2} &= -i 2 \pi \operatorname*{Res}_{z=-i} \frac{e^{i k z}}{(1+z^2)^2}\\ &= -i 2 \pi \left [\frac{d}{dz} \frac{e^{i k z}}{(z-i)^2} \right ]_{z=-i}\\ &= -i 2 \pi \left [\frac{i k\, e^{i k z}}{(z-i)^2} - \frac{2 e^{i k z}}{(z-i)^3} \right ]_{z=-i}\\ &= \frac{\pi}{2} (-k+1) e^{k}\end{align}$$
Therefore,
$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^2} = \frac{\pi}{2} (1+|k|) e^{-|k|}$$
Method 2: Convolution
Knowing that the FT of $1/(1+x^2)$ is $\pi \, e^{-|k|}$, we may use the convolution theorem to deduce that
$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{(1+x^2)^2} = \frac{\pi^2}{2 \pi} \int_{-\infty}^{\infty} dk' e^{-|k'|} \, e^{-|k-k'|}$$
Again, the way we evaluate the integral on the RHS depends on the sign of $k$. For $k \gt 0$, this integral is
$$\frac{\pi}{2} \int_{-\infty}^0 dk' \, e^{k'} \, e^{-(k-k')} + \frac{\pi}{2} \int_{0}^k dk' \, e^{-k'} \, e^{-(k-k')}+ \frac{\pi}{2} \int_{k}^{\infty} dk' \, e^{-k'} \, e^{k-k'}$$
For $k \lt 0$, on the other hand, the integral is
$$\frac{\pi}{2} \int_{-\infty}^k dk' \, e^{k'} \, e^{-(k-k')} + \frac{\pi}{2} \int_{k}^0 dk' \, e^{k'} \, e^{k-k'}+ \frac{\pi}{2} \int_{0}^{\infty} dk' \, e^{-k'} \, e^{k-k'}$$
Evaluation of the above integrals reproduces the result derived above.
As already stated it depends upon how you really define the Fourier transform.
If you define it as an improper integral the question becomes non-trivial (otherwise it is just Plancherel). For $n\geq 1$ let
$$ f_n(x) = f(x)\; {\rm \bf 1}_{[-n,n]}(x) = \frac{1}{\sqrt{1+x^2}} {\rm \bf 1}_{[-n,n]}(x) $$
where ${\rm \bf 1}_{[-n,n]}$ is the indicator function for the interval $[-n,n]$. Then $f_n$ is in both $L^1\cap L^2$ and the Fourier transform
$$ \hat{f}_n(t) = \int_{-n}^n \frac{e^{itx}}{\sqrt{1+x^2}} dx $$
converges for every $t\neq 0$. This follows from partial integration:
$$ \hat{f}_n(t) = \left[ \frac{e^{itx}}{it\sqrt{1+x^2}} \right]_{-n}^n + \int_{-n}^n \frac{\sin(tx)}{t} \frac{x}{\sqrt{1+x^2}^3} dx . $$
As $n\rightarrow \infty$, the first term goes to zero and the latter converges absolutely.
Thus, you may define $\hat{f}(t)=\lim_n \hat{f}_n(t)$ for every $t\neq 0$ and the question is then if $\hat{f}\in L^2$?
The answer is yes: First note, that $f_n \rightarrow f$ in $L^2$ so $(f_n)_{n\geq 1}$ is Cauchy in $L^2$.
By Plancherel $\|\hat{f}_n-\hat{f}_m\|_2 = \|f_n-f_m\|_2$, so $(\hat{f}_n)_{n\geq 1}$ is also Cauchy in $L^2$, whence converges in $L^2$ to some function $g$.
Any convergent sequence in $L^2$ has, however, a subsequence that converges a.e. and we already know that for $t\neq 0$ (i.e. a.e.) the limit is $\hat{f}(t)$, so $\hat{f}=g$ (mod 0) is in $L^2$
Best Answer
For any $h\in L^2(\mathbb R)$, let $D(h)=\int_{\mathbb R} u^2|h(u)|^2du$.
You're looking for a lower bound for $D(f)D(\hat g)+D(\hat f )D(g)$. It's worth noting that, without any constraint on $f$ and $g$, the lower bound is $0$ since you can plug in $f=0$ or $g=0$. In fact the whole expression scales linearly with $\|f\|$ or with $\|g\|$. So let's assume that we've normalized $f$ and $g$ so that $\|f\|=\|g\|=1$.
By the Uncertainty Principle, there is a universal constant $C>0$ such that for all $h\in L^2(\mathbb R)$ with $\|h\|=1$, $$D(h)D(\hat h)\geq C$$
The exact value of $C$ depends on the normalization convention taken for the Fourier transform (it can be $\frac 1 {16\pi^2}$ or $\frac 1 {4\pi}$). The equality is attained for Gaussian functions.
With this, $$D(f)D(\hat g)+D(\hat f )D(g)\geq C\left(\frac{D(f)}{D(g)}+\frac{D(g)}{D(f)}\right)$$
The right-hand side is minimal iff $D(f)=D(g)$. In that case, the expression to minimize is equal to $2D(f)D(\hat f)$, which, by the Uncertainty Principle, is lower-bounded by $2C$, with equality attained by Gaussian functions.
We conclude that the lower bound is 2C, with equality attained when $f$ and $g$ are both Gaussian functions.