Is a bounded Operator $T\colon X\to Y$ which satisfies $\beta\lVert x\rVert\leq\lVert Tx\rVert$ for some $\beta > 0$ and all $x\in X$ an isometry?

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The following question came to when reading L.C.Evans' proof of the Lax-Milgram theorem in his well-known PDE book. However it has nothing to do with PDEs at all but is pure functional-analysis:

Let $X$, $Y$ normed spaces and $T\colon X\to Y$ a bounded linear operator. In addition to the boundedness which means \begin{equation*} \exists\alpha > 0\,\forall x\in X\colon\qquad \lVert Tx\rVert\leq \alpha\lVert x\rVert \end{equation*} we also want the operator to satisfy the following estimate: \begin{equation*} \exists\beta > 0\,\forall x\in X\colon\qquad\beta\lVert x\rVert\leq\lVert Tx\rVert. \end{equation*} The question is now: Can we conclude that $T$ is an isometry, i.e. $\lVert Tx\rVert = \lVert x\rVert$ for all $x\in X$?

If not, can we make the conclusion come true by requiring stronger assumptions, e.g. $T$ being bijective or $X$, $Y$ being Banach, etc.?