If A is an $m\times n$ matrix, show that $||A||_2 \le \sqrt{||A||_1*||A||_\infty}$ 
If $A$ is an $m\times n$ matrix, show that $$\| A \|_2 \le \sqrt{\|A\|_1 \, \| A \|_\infty}$$


I reduced this to:
$$\rho(A^TA) \le||A||1*||A||_\infty$$
I created a matrix for experimenting:
$$A = \begin{bmatrix}a&b\\c&d\end{bmatrix} \\
A^T = \begin{bmatrix}a&c\\b&d\end{bmatrix} \\
A^tA = \begin{bmatrix}a^2+b^2&ac+bd\\ac+bc&c^2+d^2\end{bmatrix}$$
I tried to calculate the eigenvalues but that got me nowhere.
$$-\lambda^2-\lambda(a^2+b^2+c^2+d^2)+(ad-bc)^2 = 0 \Rightarrow(ad-bc)^2 = \lambda(a^2+b^2+c^2+d^2)+\lambda^2$$
As for $||A||_1$ it can be either $a+c$ or $b+d$, and $||A|||_\infty$ can be either $a+b$ or $c+d$. This gives me 4 possible combinations:
$$c^2+cd+a(c+d) \\
a^2+a(b+c)+bc \\
b^2+a(b+d)+db \\
d^2+d(b+c)+bc$$
I suppose I could do:
$$\lambda = \frac{(a^2+b^2+c^2+d^2)\pm \sqrt{(a^2+b^2+c^2+d^2)^2+4*(ad-bc)^2}}{-2}$$
Since that root will always be bigger than $(a^2+b^2+c^2+d^2)$, $\lambda$ will be either negative (because of the -2), in which case it's "proven" because the right hand side of the initial equation is always positive, or... it will be so big that the denominator will approach zero from the negative side, when the sign is negative, because you'll have $(a^2+b^2+c^2+d^2)-((a^2+b^2+c^2+d^2)+ something)$. In that case, the result will be a very small negative value, divided by -2, which is a very small positive value.
Going from there it's "intuitive" that the equation is correct but not rigorous, I suppose.
Help?

The conventional proof is
$$
\|A\|_2^2=\rho(A^\ast A)\le\|A^\ast A\|_\infty\le\|A^\ast\|_\infty\|A\|_\infty=\|A\|_1\|A\|_\infty,
$$
where the first inequality is due to the fact that the spectral radius is the infimum of all submulticative norms, and the second inequality is due to the submultiplicativity of the induced maximum norm.
