It is known that $(\sum_{i=1}^na_i)^p \leq 2^p\sum_{i=1}^na^p_i$ for $p\geq 1$ with $a_1, \ldots, a_n \geq 0,$ but:
Question1: How do I prove $$\Big(\sum_{i=1}^na_i\Big)^\nu \leq \;\sum_{i=1}^na^\nu_i$$ with $\nu \in (0,1]$ and $a_1, \ldots, a_n \geq 0\,?$
Question2: Is there even a constant $c_\nu \in (0,1),$ s.t.: $$\Big(\sum_{i=1}^na_i\Big)^\nu \leq \;c_\nu\sum_{i=1}^na^\nu_i\;\;\;?$$
Without loss of generality, suppose that $a_1\leq a_2\leq \ldots \leq a_n$. If $a_n=0$, then all $a_i$'s are $0$ and the first inequality holds. From now on, suppose that $a_n>0$.
Note that $$\left(\sum_{i=1}^n\,a_i\right)^\nu=a_n^\nu\,\left(1+\sum_{i=1}^{n-1}\,\frac{a_i}{a_n}\right)^\nu\,.$$ Using Bernoulli's Inequality, we get $$\left(\sum_{i=1}^n\,a_i\right)^\nu\leq a_n^\nu\,\left(1+\nu\,\sum_{i=1}^{n-1}\,\frac{a_i}{a_n}\right)\,.$$ Clearly, $\displaystyle a_n^\nu\left(\frac{a_i}{a_n}\right)=\frac{a_i}{a_n^{1-\nu}}\leq a_i^\nu$ for $i=1,2,\ldots,n-1$. Thence, $$\left(\sum_{i=1}^n\,a_i\right)^\nu\leq a_n^\nu+\nu\,\sum_{i=1}^{n-1}\,a_i^\nu\leq \sum_{i=1}^n\,a_i^\nu\,.$$ The equality occurs if and only if $\nu=1$ or $a_1=a_2=\ldots=a_{n-1}=0$.
To answer the second question, take $a_1=a_2=\ldots=a_{n-1}=0$. Then, the (second) inequality reads $a_n^\nu\leq c_\nu\,a_n^\nu$ for all $a_n\geq 0$. Ergo, $c_\nu \geq 1$. That is, there is no such constant $c_\nu\in(0,1)$. However, one can prove, using the Power-Mean Inequality, that $$\sum_{i=1}^n\,a_i^\nu\leq n^{1-\nu}\,\left(\sum_{i=1}^n\,a_i\right)^\nu\,,$$ in which the equality happens iff $\nu=1$ or $a_1=a_2=\ldots=a_n$.