= Solution
If $T$ is invertible, the <bounded inverse theorem> makes $T^{-1}$ bounded, and
$$
\|Tx\|\geq\|T^{-1}\|^{-1}\|x\|.
$$
Moreover $T^*$ is invertible with inverse $(T^{-1})^*$, so the same argument shows that $T^*$ is bounded below.
Conversely, suppose
$$
\|Tx\|\geq c\|x\|,
\qquad
\|T^*x\|\geq d\|x\|
$$
for some $c,d>0$. The first bound makes $T$ injective. It also makes $\operatorname{ran}T$ closed: if $Tx_n$ is Cauchy, then $x_n$ is Cauchy, and continuity gives its limit in the range. The second bound gives $\ker T^*=\{0\}$. By <image-kernel orthogonality for an adjoint>,
$$
(\operatorname{ran}T)^\perp=\ker T^*=\{0\},
$$
so the range is dense. Being both closed and dense, it equals $H$. Thus $T$ is bijective, and the first lower bound gives $\|T^{-1}\|\leq c^{-1}$. Therefore
$$
\boxed{T\text{ is invertible}
\quad\Longleftrightarrow\quad
T\text{ and }T^*\text{ are bounded below}.}
$$
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