= Solution
The lower bound in (ii) makes the <image> of $I+T$ closed: if $(I+T)x_n$ converges, the bound makes $(x_n)$ Cauchy; completeness gives its limit and hence a <image> preimage. The same bound applies to $I+T^*$ by (iii), so its <kernel> is zero. The <orthogonal complement> identity
$$
\operatorname{ran}(I+T)^\perp=\ker(I+T^*)=\{0\}
$$
shows that the <image> is dense, using the <orthogonal decomposition by a closed subspace> for its <closure>. Being both closed and dense, it equals $X$. Thus $I+T$ is onto and bounded below, and by part 3(a)(iii) it has a bounded inverse. Moreover
$$
\boxed{\|(I+T)^{-1}\|\leq1.}
$$
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