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The problem comes from Naylor's Linear Operator Theory Section 5.23 Problem 14

Let $T$ be a normal operator on Hilbert space $H$ and let $T = A + iB$ be the Cartesian decomposition of $T$. Show that $$\mathrm{max}(\left\lVert A\right\rVert^2,\left\lVert B\right\rVert^2) \leq\left\lVert T\right\rVert^2 \leq \left\lVert A\right\rVert^2+\left\lVert B\right\rVert^2$$

I've shown that Cartesian decomposition of $T$ is unique, $T^*=A-iB$, and that $A$ and $B$ commutes.

For the $\left\lVert T\right\rVert^2 \leq \left\lVert A\right\rVert^2+\left\lVert B\right\rVert^2$ part, I used the fact that $T$ is normal and use that $\left\lVert T\right\rVert^2 =\left\lVert T T^*\right\rVert = \left\lVert T^2 \right\rVert$.

However the $\mathrm{max}(\left\lVert A\right\rVert^2,\left\lVert B\right\rVert^2) \leq\left\lVert T\right\rVert^2$ part is giving me trouble. At first, I made a series of inequalities to build up to$\left\lVert T\right\rVert^2$, but no luck.

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From the properties of the decomposition, one gets $$ \|Tx\|^2 =\|Ax\|^2 + \|Bx\|^2, $$ hence $$ \max( \|Ax\|^2, \|Bx\|^2 ) \le \|Tx\|^2 \le \|T\|^2 \|x\|^2. $$ Now take the supremum over $x$ with $\|x\|\le 1$.

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