Extending $\|H^{\frac{1}{2}}XK^{\frac{1}{2}}\|\leq\frac{1}{2}\|HX+XK\|$ from matrices to operators I saw in some literature that many author works in finite dimensional (matrix) is because it can be extended into infinite dimensional (operator). The case is as follows:
If the following inequality 
$$\|H^{\frac{1}{2}}XK^{\frac{1}{2}}\|\leq\frac{1}{2}\|HX+XK\|$$ holds for any $n\in\mathbb{N}$ and for any $H,K,X\in M_{n}(\mathbb{C})$ with $H,K>0,$ then so is for any $H,K,X\in\mathcal{B}(\mathcal{H})$ with $H,K>0.$
Here $M_{n}(\mathbb{C})$ denotes the collection of $n\times n$ matrix with complex entries, $H>0$ denotes positive definite matrix (operator), $\mathcal{B}(\mathcal{H})$ denotes the collection of bounded linear operator on Hilbert space $\mathcal{H}.$
What I'm thinking is by involving (finite dimensional) projection and take the limit.
$$\lim_{n\rightarrow\infty}\|P_{n}H^{\frac{1}{2}}XK^{\frac{1}{2}}P_{n}\|\leq\frac{1}{2}\lim_{n\rightarrow\infty}\|P_{n}(HX+XK)P_{n}\|$$
where $\lim_{n\rightarrow\infty} P_{n}=I,$ ($I\in\mathcal{B}(\mathcal{H})$). 
But I'm not sure, whether I'm on the right track or not. 
Any help would be appreciated. Thank you

I dont'think you are on the right track. The limit $\lim_nP_n=I$ occurs in the strong operator topology (and other weak topologies on $B(H)$) but not in the norm topology. 
If you are looking for the particular inequality mentioned in your question, the usual proof works the same for operators, you gain nothing by going to matrices first. 
If your question is general, I don't think you can expect to have a general method that allows you to prove that any norm inequality for matrices holds for arbitrary operators. 
