**Method 3: Lagrange multipliers.**

Let $d>0$ and $U=\{(a,b);\;a>0,\;b>0\}$. Define $g:U\to\mathbb{R}$ and $\varphi:U\to \mathbb{R}$ by $g(a,b)=\left(\frac{1}{a+d}+1\right)\left(\frac{1}{b+d}+1\right)$ and $\varphi(a,b)=a+b$.
Then,
$$\nabla g(a,b)=-\left(\tfrac{1}{A^2}\left(\tfrac{1}{B}+1\right),\tfrac{1}{B^2}\left(\tfrac{1}{A}+1\right)\right)$$
where $A=a+d$ and $B=b+d$, and $\nabla\varphi(a,b)=(1,1)$.

It follows from the Lagrange Multipliers Method that:

$$\begin{align}(a,b)\in \varphi^{-1}(\lambda)\text{ is a critical point of } g\quad&\Longleftrightarrow\quad\nabla g(a,b)=\gamma \nabla \varphi(a,b) \text{ for some } \gamma\in\mathbb{R}\\\\
&\Longleftrightarrow \qquad\frac{1}{A^2}\left(\frac{1}{B}+1\right)
=\frac{1}{B^2}\left(\frac{1}{A}+1\right)\\\\
&\Longleftrightarrow\qquad B+B^2=A+A^2\\\\
&\Longleftrightarrow\qquad B-A=A^2-B^2=-(A+B)(B-A)\\\\
&\Longleftrightarrow\qquad A=B\\\\
&\Longleftrightarrow\qquad a=b=\frac{\lambda}{2}\end{align}$$

So, $q=(\frac{\lambda}{2},\frac{\lambda}{2})$ is the unique critical point of $g$ in $\varphi^{-1}(\lambda)$.

It is possible to prove that $g$ has indeed a minimum at $q$ (see the argument here).

Therefore, for all $(a,b)\in\varphi^{-1}(\lambda)$ we have
$$\left(\frac{1}{a+d}+1\right)\left(\frac{1}{b+d}+1\right)=g(a,b)\geq g\left(\tfrac{\lambda}{2},\tfrac{\lambda}{2}\right)=\left(\frac{1}{\tfrac{\lambda}{2}+d}+1\right)\left(\frac{1}{\tfrac{\lambda}{2}+d}+1\right)$$

Letting $d\to 0$ we conclude that
$$f(a,b)^2=\left(\frac{1}{a}+1\right)\left(\frac{1}{b}+1\right)\geq \left(\frac{1}{\tfrac{\lambda}{2}}+1\right)\left(\frac{1}{\tfrac{\lambda}{2}}+1\right)=f\left(\tfrac{\lambda}{2},\tfrac{\lambda}{2}\right)^2$$
and thus the the minimum is attained at $(\tfrac{\lambda}{2},\tfrac{\lambda}{2})$.