Inequality involving two normal operators Let $E$ be a complex Hilbert space, with inner product $\langle\cdot\;, \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{L}(E)$ the algebra of all bounded linear operators from $E$ to $E$. 

Let $A_1, A_2\in \mathcal{L}(E)$ be two normal operators such that $A_1A_2=A_2A_1$.
  I need some help to know how can I prove this inequality:
  $$\displaystyle\sup_{\|f\|=1}\bigg(|\langle A_1f\;,\;f\rangle|^2+|\langle A_2f\;,\;f\rangle|^2\bigg)^{1/2}\geq\displaystyle\sup_{\|f\|=1}\bigg(\|A_1f\|^2+\|A_2f\|^2\bigg)^{\frac{1}{2}}.$$

I am trying to solve it, but I did not reach to any answer, I would be so glad if anyone could help me on that.
Thank you everyone !!!
 A: Using the spectral decomposition, the statement is equivalent to
$$
\sup_{\|f\|=1} \left( \left|\int \phi_1 |f|^2\right|^2 + \left|\int \phi_2 |f|^2\right|^2 \right)
\ge \sup_{\|f\|=1} \left( \int |\phi_1|^2 |f|^2 + \int |\phi_2|^2 |f|^2\right).
$$
It is easy to see, that the right-hand side is equal to $\|\phi_1^2 + \phi_2^2\|_{L^\infty}$ (just set $f$ to be a multiple of the characteristic function of $\{x: |\phi_1|^2 + |\phi_2|^2 \ge \||\phi_1|^2 + |\phi_2|^2\|_{L^\infty}-\epsilon\}$).
The same idea can be applied for the left-hand side in the real case.
Let me assume first that $\phi_1,\phi_2$ are real-valued and non-negative(or equivalently $A_1,A_2$ self-adjoint and positive semidefinite).
Let $s,t\ge0$ be given.
Set $S:=\{x: \ \phi_1\ge s, \phi_2\ge t\}$. 
Suppose $S$ has positive measure.
Set $f:=\chi_S \|\chi_S\|_{L^2}^{-1}$. Then
$$
(\int \phi_1 f^2)^2 + (\int \phi_2 f^2)^2 \ge s^2 + t^2.
$$
Set $M:=\||\phi_1|^2 + |\phi_2|^2\|_{L^\infty}$.
Let $\epsilon\in(0,M)$ be given. Then there are nonnegative $s,t$ such that $s^2+t^2\ge M-\epsilon$ and $S:=\{x: \ \phi_1\ge s, \phi_2\ge t\}$ has positive measure.
This shows that the left-hand side is $\ge \||\phi_1|^2 + |\phi_2|^2\|_{L^\infty}$. This proves the claim for non-negative $\phi_i$. 
For arbitrary  $\phi_i$, one can argue similarly: There are numbers $s,t\ge0$ 
with $s^2+t^2\ge M-\epsilon$ and complex unit vectors $u_1,u_2$ such the set 
$$S:=\{x: \ \Re(u_1\phi_1)\ge s, \Re(u_2\phi_2)\ge t\}$$ 
has positive measure. Then for $f:=\chi_S \|\chi_S\|_{L^2}^{-1}$ we get
$$
\left|\int \phi_1 |f|^2\right|^2 + \left|\int \phi_2 |f|^2\right|^2
\ge \left(\Re u_1\int \phi_1 |f|^2\right)^2 + \left(\Re u_2\int \phi_2 |f|^2\right)^2\ge s^2 + t^2.
$$
