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Given that $A$ is symmetric, for any norm $\|\cdot\|_*$: $$\|A\|_2 \le \|A\|_*$$

Why is it so? I think it has something to do with the fact that $\rho(A) \le \|A\|$ (Where $\rho (A)$ is the spectral radius of $A$)

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Your intuition is correct. If $A$ is symmetric, it is orthogonally diagonalisable and hence its singular values are just the absolute values of its eigenvalues. Hence $\|A\|_2=\rho(A)$.

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    Oh nice! Thank you @user1551! – 2017-02-25
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    This works for sure if $\|\cdot\|_*$ is an induced norm, but what if it is not? – 2017-02-25
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    @JackD'Aurizio I suppose the OP is talking about *submultiplicative* matrix norms (including, but not limited to, induced norms), otherwise the statement is obviously wrong because any vector norm on the matrix space can be scaled by an arbitrarily small positive factor to form another vector norm. Now, if the norm is submultiplicative, $\rho(A)\le\|A\|_\ast$ follows directly from Gelfand's formula. – 2017-02-25