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$|||A||| = \max \frac{\left^{\frac{1}{2}}}{||x||}$

But how can I relate $$\left$$ only to $|||A|||$ or some other matrix norm?

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    There isn't a general relation between the two. The matrix norm does bound the expression you have, however. – 2017-02-11
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    @paul why it is bounded by the norm? Could you explain more? – 2017-02-11

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It's Cauchy Schwarz: $$ |\langle x|A|x\rangle|\leq \|x\|\,\|Ax\|. $$ And $$ \|Ax\|^2=\langle x|A^\dagger A|x\rangle\leq \|A\|^2\,\|x\|^2. $$ So $$ |\langle x|A|x\rangle|\leq\|A\|\,\|x\|^2. $$

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    But A need not be self adjoint. Does it hold of A is not self adjoint? – 2017-02-11
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    I'm not requiring $A$ to be selfadjoint at any step. – 2017-02-11
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    @Omnomnomnom: I'm using the OP's definition. – 2017-02-11
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    @MartinArgerami I was misreading it the whole time! Sorry. – 2017-02-11
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    No worries, I do it more often than you do. – 2017-02-11
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    But since whether or not A is self -adjoint or not $|\langle x|A|x\rangle|\leq\|A\|\,\|x\|^2$. Does A being self-adjoint result in a tighter bound as there is no need for Cauchy Schwarz if A is self-adjoint? – 2017-02-11
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    @peter: you need CS, whether $A$ is selfadjoint or not. It doesn't make a difference. And, for any $A$, whether selfadjoint or not, you can always find $x$ such that the two sides of the inequality are arbitrarily close, so the inequality is as sharp as it can be. – 2017-02-12