a question about contractions on Hilbert spaces Let $\cal{H}$ be a Hilbert space, $T_1,T_2\in\cal{B(H)}$, 


$\|T_1(h_1)+T_2(h_2)\|^2\leq\|h_1\|^2+\|h_2\|^2$ for all $h_1,h_2\in\cal{H}$.
$T_1T^\ast_1+T_2T^\ast_2\leq I$.


Then can we verify that 1 holds if and only if 2 holds ?
Consider the operator $T:\mathcal H\oplus \mathcal H\to\mathcal H$ defined by 
$$T(h_1\oplus h_2):= T_1(h_1)+T_2(h_2)\, . $$

Then (1) says exactly that $\Vert T\Vert\leq 1$. Since $\Vert T\Vert=\Vert T^*\Vert$, this is equivalent to the condition $\Vert T^*\Vert\leq 1$, which is again equivalent to 
$$TT^*\leq I\, .$$

Now, you will easily verify that $T^*:\mathcal H\to \mathcal H\oplus \mathcal H$ is given by the formula
$$T^*(u)=T_1^*(u)\oplus T_2^*(u)\, . $$

So we have $T(T^*(u))=T_1T_1^*(u)+T_2T_2^*(u)$, i.e.
$$TT^*=T_1T_1^*+T_2T_2^*\, .$$

Hence, (1) is equivalent to (2).
As Paul remarked above- this is not true. Take $h_1$ and $h_2$ to be two orthogonal vectors in $\mathcal{H}$, and $T_1=T_2=I$. Then, your condition 1) holds, but condition 2) doesn't.