I need to prove that for $T' \in \mathcal{B}(X^*)$, where $T'$ is invertible that $$\forall x \in X \Vert Tx\Vert ≥\Vert(T')^{-1}\Vert^{-1} \Vert x\Vert$$ Note that I don't yet know that $T$ is itself invertible; the next part of the question is to prove that.
The closest I could get was to show $$\Vert T\Vert \Vert x\Vert≥\Vert(T')^{-1}\Vert^{-1} \Vert x\Vert$$ But I didn't use any corollary of the Hahn-Banach Theorem to do this, which is what the question hints to do. Any help would be appreciated!
Let $x\in X$. $$\|Tx\|=\sup_{x^*\in X^*,\|x^*\|≤1} \|x^*(T(x))\|=\sup_{x^*\in X^*,\|x^*\|≤1} \|(T'x^*)(x)\|$$ Now $T'$ is an invertible map, and by Hahn Banach there exists an $x^*$ with $\|x^*\|=1$ so that $x^*(x)=\|x\|$. Take $\frac{T'^{-1}x^*}{\|T'^{-1}x^*\|}$ as a special point over which the supremum is being taken. Then you get: $$\|Tx\|≥\frac{\|T'(T'^{-1}x^*)\,(x)\|}{\|T'^{-1}x^*\|}=\frac{x^*(x)}{\|T'^{-1}x^*\|}≥\frac{\|x\|}{\|T'^{-1}\|\,\|x^*\|}=\frac{\|x\|}{\|T'^{-1}\|}$$