Let $a=\tan\alpha$ and $b=\tan\beta$ where $\alpha, \beta\in(0, \frac{\pi}{4}]$. Then $$\begin{array}{lrcl} &\frac{1}{\sqrt{a^2 + 1}}+\frac{1}{\sqrt{b^2 +1}}&\le&\frac{2}{\sqrt{1+ab}}\\ \Leftrightarrow&\frac{1}{\sqrt{\tan^2\alpha+1}}+\frac{1}{\sqrt{\tan^2\beta+1}}&\le&\frac{2}{\sqrt{1+\tan\alpha\tan\beta}}\\ \Leftrightarrow&\cos\alpha+\cos\beta&\le&2\sqrt{\frac{\cos\alpha\cos\beta}{\cos\alpha\cos\beta+\sin\alpha\sin\beta}}\\ \Leftrightarrow&\cos\alpha+\cos\beta&\le&2\sqrt{\frac{\cos\alpha\cos\beta}{\cos(\alpha-\beta)}} \end{array}$$ Because $\alpha, \beta\in(0, \frac{\pi}{4}]$, it must hold that $$0 < \cos\alpha, \cos\beta, \cos(\alpha-\beta) < 1$$ Hence, the last inequality is equivalent to its square, i.e.: $$\begin{array}{lrcl} &(\cos\alpha + \cos\beta)^2&\le&4\cdot\frac{\cos\alpha\cos\beta}{\cos(\alpha-\beta)})\\ \Leftrightarrow&\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta&\le&\frac{4\cos\alpha\cos\beta}{\cos(\alpha-\beta)}\\ \Leftrightarrow&\cos(\alpha-\beta)(\cos^2\alpha+\cos^2\beta)&\le&(4-2\cos(\alpha-\beta))\cos\alpha\cos\beta \end{array}$$ Because $0 < \cos(\alpha-\beta) < 1$, therefore it is sufficient to show $$\begin{array}{lrcl} &\cos(\alpha-\beta)(\cos^2\alpha+\cos^2\beta)&\le&(4-2)\cos\alpha\cos\beta\\ \Leftrightarrow&\cos(\alpha-\beta)(\cos^2\alpha+\cos^2\beta)&\le&2\cos\alpha\cos\beta\\ \Leftrightarrow&\cos(\alpha-\beta)\Big(\frac{1+\cos 2\alpha}{2}+\frac{1+\cos 2\beta}{2}\Big)&\le&2\cos\alpha\cos\beta\\ \Leftrightarrow&\cos(\alpha-\beta)(\cos 2\alpha + \cos 2\beta +2)&\ge&4\cos\alpha\cos\beta \end{array}$$ Applying sum-to-product transformation on the left gives $$\cos(\alpha-\beta)(2\cos(\alpha+\beta)\cos(\alpha-\beta)+2)$$ Applying product-to-sum transformation on the right gives $$2(\cos(\alpha+\beta)+\cos(\alpha-\beta))$$ Setting these back and rearranging yields $$\begin{array}{lrcl} \Leftrightarrow & \cos^2(\alpha-\beta)\cos(\alpha+\beta) &\le&\cos(\alpha+\beta) \end{array} $$ This clearly holds because $\cos^2(\alpha-\beta) \le 1$ and $$0 < \alpha, \beta \le \frac{\pi}{4}\implies 0 < \alpha + \beta \le \frac{\pi}{2}\implies 0 \le \cos(\alpha + \beta) < 1$$