It is evident that the inequality under consideration becomes an equality when $a=0$, $b=0$ or $a=b$. To prove that otherwise the strong inequality holds, it suffices to deal with the case $a > b > 0$ and (after removing the fractions) to show that
$$a\sqrt{a^2+1} + b\sqrt{b^2+1} > (a+b)\sqrt{a^2+1}$$
Distributing the right-hand side and regrouping the terms we get
$$a\sqrt{a^2+1}(\sqrt{ab+1} - \sqrt{b^2+1}) > b\sqrt{b^2+1}(\sqrt{a^2+1} - \sqrt{ab+1})$$
Multiplying the differences of the square roots by their sums as the denominators of new installed fractions, we obtain
$$a\sqrt{a^2+1} \cdot \frac{b(a-b)}{\sqrt{ab+1} + \sqrt{b^2+1}} > b\sqrt{b^2+1} \cdot \frac{a(a-b)}{\sqrt{a^2+1} + \sqrt{ab+1}}$$
Dividing both sides by the positive number $ab(a-b)$ and removing the fractions again, we finally arrive at an equivalent inequality
$$\sqrt{a^2+1}(\sqrt{a^2+1} + \sqrt{ab+1}) > \sqrt{b^2+1}(\sqrt{b^2+1} + \sqrt{ab+1})$$
which easily follows by an easy comparison of the both sides "term by term" (because our assumption $a > b$ implies that $\sqrt{a^2+1} > \sqrt{b^2+1}$). This completes the proof of the given inequality. As we have shown, the only cases of the equality are $a = 0$, $b = 0$ and $a = b$.

Another solution:
We exclude the cases $a = 0$ and $b = 0$ (when the inequality becomes an equality) from our considerations. Let us apply a Cauchy-Schwarz inequality in the form
$$\left(\frac{a}{u} + \frac{b}{v}\right) (au + bv) \geq (a+b)^2,$$
with positive coefficients $u = \sqrt{b^2 + 1}$ and $v = \sqrt{a^2 + 1}$:
$$\left( \frac{a}{\sqrt{b^2+1}} + \frac{b}{\sqrt{a^2+1}} \right) \left( a\sqrt{b^2+1} + b\sqrt{a^2+1} \right) \ge (a+b)^2. \quad (1)$$
Another Cauchy-Schwarz inequality yields an upper bound for the second factor from the-left hand side of (1):
$$\begin{align*} a\sqrt{b^2+1} + b\sqrt{a^2+1} &= \sqrt{a}\sqrt{ab^2+a} + \sqrt{b}\sqrt{a^2b+b} \le \\ &\le \sqrt{a+b}\sqrt{ab^2+a+a^2b+b} = \sqrt{a+b}\sqrt{(a+b)(ab+1)} = (a+b)\sqrt{ab+1}. \end{align*}$$
Consequently, the first factor in (1) has a lower bound
$$\frac{a}{\sqrt{b^2+1}} + \frac{b}{\sqrt{a^2+1}} \ge \frac{(a+b)^2}{a\sqrt{b^2+1} + b\sqrt{a^2+1}} \ge \frac{a+b}{\sqrt{ab+1}},$$
which is the desired inequality. Since (1) becomes an equality if and only if the positive coefficients $u$ and $v$ are the same, i.e. $\sqrt{b^2+1} = \sqrt{a^2+1}$ in our situation, the equality $a = b$ is the third (and last) case (next to $a = 0$ and $b = 0$ from the introductory sentence) when the proven inequality holds as an equality.

Another solution:
Let us exclude the obvious cases $a=0, b=0, a=b$ and let us transform the (strong) inequality under consideration into the following equivalent form:
$$\frac{a}{a+b} \cdot \frac{1}{\sqrt{b^2+1}} + \frac{b}{a+b} \cdot \frac{1}{\sqrt{a^2+1}} > \frac{1}{\sqrt{ab+1}} \quad (2)$$
The last left-hand side can be read as that of the (strong) Jensen inequality
$$\mathrm{pf}(\alpha) + \mathrm{qf}(\beta) > f(\rho\alpha + q\beta), \quad (3)$$
with positive coefficients $p = a/(a+b)$ and $q = b/(a+b)$ (which satisfy $p+q=1$ as required), applied to the function $f(x) = 1/\sqrt{x}$ at the points $\alpha = b^2+1$ and $\beta = a^2+1$. Since the function $f$ is strictly convex on the interval $(0, +\infty)$ and since the points $\alpha$ and $\beta$ are assumed to be distinct, the Jensen inequality (3) holds.
It remains to verify that also the right-hand sides of (2) and (3) are identical. This is easy:
$$\begin{align*} f(p\alpha + q\beta) &= f\left(\frac{a}{a+b}(b^2+1) + \frac{b}{a+b}(a^2+1)\right) = \\ &= f\left(\frac{a + ab^2 + b + a^2b}{a+b}\right) = f(ab+1) = \frac{1}{\sqrt{ab+1}}. \end{align*}$$