An inequality in probability theory I am trying to understand a paper but am stuck at some point. However I could understand it if I proved the following result, but I am not sure whether it acutally is true.
$\mathbf{Conjecture:}$ let $Z_1.\, \dots, Z_n $ be iid $\sim Z \in L^2$, with $\mathbb{E}Z=0$. Then
$$
\mathbb{E}[\sqrt{\sum_{i=1}^n Z_i^2}] \leq \sup_{\lambda\in \mathbb{R}^n : ||\lambda||_2 = 1} \mathbb{E}[|\sum_{i=1}^n \lambda_i Z_i|]
$$
I think something like this is being us in the paper, but I cannot see whether it is true. Any help?
 A: This is false.  Take $n=2$ and take $Z_1, Z_2$ i.i.d. Gaussian with mean 0 and variance 1. By symmetry, the following expression is the same for all $(\lambda_1, \lambda_2) \in \mathbb{R}^2$ that satisfy $\lambda_1^2 + \lambda_2^2=1$:
$$ E[|\lambda_1 Z_1 + \lambda_2Z_2|] $$
Thus this is the same for $(\lambda_1, \lambda_2) = (1,0)$ and
$$ E[|\lambda_1 Z_1 + \lambda_2Z_2|] = E[|Z_1|] = \sqrt{\frac{2}{\pi}}\approx 0.79788456  \quad \mbox{(whenever $\lambda_1^2+\lambda_2^2=1$)}$$
On the other hand
$$ E\left[\sqrt{Z_1^2 + Z_2^2} \: \right] = \sqrt{\frac{\pi}{2}} \approx 1.2533141$$

Edit:
To give one proof of the symmetry observation: Consider the rotation matrix
$$ A = \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}$$
Define
$$ \begin{bmatrix}X_1\\ X_2\end{bmatrix} = A\begin{bmatrix}Z_1\\Z_2\end{bmatrix}$$
Then $X_1, X_2$ are again i.i.d. Guassian mean 0, variance 1.  So for any measurable function $f(x)$ (including $f(x)=|x|$) and any constant $\theta \in \mathbb{R}$:
\begin{align}
E\left[f\left([\cos(\theta) \: \sin(\theta)] \cdot \begin{bmatrix}Z_1\\Z_2\end{bmatrix}\right)\right] &= E\left[f\left([\cos(\theta) \: \sin(\theta)] \cdot A\begin{bmatrix}Z_1\\Z_2\end{bmatrix}\right)\right]\\
&=E\left[f\left([1 \: \: 0] \begin{bmatrix}Z_1\\Z_2\end{bmatrix}\right)\right]\\
&=E[f(Z_1)]
\end{align}
where the first equality uses the fact that random vector $[Z_1;Z_2]$ has the same distribution as random vector $A[Z_1;Z_2]$; the second equality uses the fact that
$$  [\cos(\theta) \: \sin(\theta)] \cdot A = [1 \: \: 0]$$
