Consider $\frac{1}{T}\sum_{t=1}^{T}\max\{ 0,a_t\}$. Can we say whether this is greater or equal then $\max\{ 0,\frac{1}{T}\sum_{t=1}^{T}a_t\}$?
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The Jensen inequality states that, if $f(x)$ is convex, then
$$ \frac{\sum_{t=1}^{T}f(x_t)}{T} \ge f\left(\frac{\sum_{t=1}^{T}x_t}{T}\right) $$
Now, $f(x)=\max(0,x)$ is convex. Hence
$$\frac{1}{T}\sum_{t=1}^T{\max\{0,x_t\}} \ge \max\left\{0,\frac{1}{T}\sum_{t=1}^T{x_t}\right\} $$
leonbloy
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Since $a_t\leq \max\{0,a_t\}$ for all $t$, it follows that $\frac{1}{T}\sum_{t=1}^T{a_t}\leq \frac{1}{T}\sum_{t=1}^T{\max\{0,a_t\}}$. Now, the RHS is always non-negative; if the LHS is non-negative, then it is equal to the max of $0$ and itself, and otherwise it is less than 0. In either case, $$\max\left\{0,\frac{1}{T}\sum_{t=1}^T{a_t}\right\}\leq \frac{1}{T}\sum_{t=1}^T{\max\{0,a_t\}}$$
Hayden
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