Equivalence of operators let $T$ and $S$ be positive definite (thus self-adjoint) operators on a Hilbert space.
I am wondering whether we have equivalence of operators 
$$ c(T+S) \le \sqrt{T^2+S^2} \le C(T+S)$$ 
for some appropriate $c$ and $C$.
This is somewhat motivated by the equivalence of all $p$ norms on a Hilbert space. 
The first inequality is trivial 
since $T^2\le T^2+S^2$ and $S^2 \le T^2+S^2$ we conclude since the square root is operator monotone that 
$$ \frac{T+S}{2} \le \sqrt{T^2+S^2}.$$
However, I do not find it very obvious whether 
$$\sqrt{T^2+S^2} \le C(T+S)$$
is possible?
 A: There is no real $C>0$ such that 
$$\sqrt{T^2+S^2} \le C(T+S) \tag{1}$$
holds for all positive definite (self-adjoint) operators on a Hilbert space of any dimension $\ge2$. 
Indeed, take any any real $C>0$ and identify $T$ and $S$ with the $2\times2$ matrices 
$$T:=\left(
\begin{array}{cc}
 1 & 0 \\
 0 & s^5 \\
\end{array}
\right),\quad 
S:=\left(
\begin{array}{cc}
 \dfrac{s^2}{4} & \dfrac{s^3}{8} \\
 \dfrac{s^3}{8} & \dfrac{s^6}{64}+\dfrac{s^4}{16} \\
\end{array}
\right), 
$$
where 
$$s:=1/C. 
$$
If (1) holds for some real $C>0$, then it holds for any large enough $C>0$, because $T+S\ge0$. 
Now, letting $C\to\infty$ (so that $s\downarrow0$), we have 
$$\det\big(C(T+S)-\sqrt{T^2+S^2}\,\big)=-\frac{s^2}{16}+O\left(s^3\right)<0
$$
for all large enough $C>0$. 
Thus indeed, there is no real $C>0$ such that (1)
holds for all positive definite operators on a Hilbert space of any dimension $\ge2$.

Here is an image of the Mathematica notebook with the relevant calculations: 

A: A slightly simplified version of Iosif Pinelis' counterexample: for any $n\ge1$ let
$$S:=\left[ \begin {array}{cc} 1& \sqrt{n-1}\\   \sqrt{n-1}&n\end {array} \right],\qquad T:=\left[ \begin {array}{cc} 1&- \sqrt{n-1}\\  - \sqrt{n-1}&n\end {array} \right] $$
Then $$S+T=2\,  {\rm diag}\big( 1,\, n\big)$$ while $$ S^2+T^2 =2\, {\rm diag}\big(  n,\, n^2+n-1\big),$$
so
$$ \sqrt{S^2+T^2 } =\sqrt{2}\, {\rm diag}\big( \sqrt{n  }\,,\, \sqrt{n^2+n-1}\big),$$ 
and comparing the $(1,1)$ entries, we have that $C$ must be at least $\sqrt\frac{{n}}{2}$ in order that $$ (S^2+T^2)^{1/2}\le C(S+T).$$ 
A: An additional remark (too long for a comment) that might be of interest:
Given the finite-dimensional counterexamples in the answers by Iosif Pinelis and Pietro Majer it seems worthwhile to note that we can use those examples to construct a counterexample in infinite dimension which is "stronger" in the sense that, for fixed $T$ and $S$, there does not exist any $C \ge 0$ the satisfies the required inequality, but "weaker" in the sense that the operators involved are only positive semi-definite:
Example. There exist positive semi-definite operators $T$ and $S$ on the separable complex Hilbert space such that
$$
\sqrt{T^2 + S^2} \le C (S+T) \qquad (*)
$$
does not hold for any $C \ge 0$.
Indeed, for each $n \in \mathbb{N}$ there exist, by the other answers, operators $T_n,S_n$ on $\mathbb{C}^2$ such that $\sqrt{T_n^2 + S_n^2} \not\le n (S_n+T_n)$. Since $(*)$ is invariant under multiplication of both $T$ and $S$ with the same positive number, we may assume that $\|T_n\| \le 1$ and $\|S_n\| \le 1$ for each $n$.
Let us then consider the operators $T = \oplus_{n \in \mathbb{N}} T_n$ and $S = \oplus_{n \in \mathbb{N}} S_n$ on the Hilbert space $\ell^2(\mathbb{N}; \mathbb{C}^2)$; it follows that those two operators do not satisfy $(*)$ for any $C \ge 0$.
