Prove that $||(\sum^k_{i=1}|f_i|)^{\frac{1}{2}}||_p \leq (\sum^k_i ||f_i||^2_p)^{\frac{1}{2}}$ In one popular Chinese Real Analysis textbook, I met a problem stated as the title: Prove the following inequality
\begin{equation}
\left|\left|\left(\sum^k_{i=1}|f_i|\right)^{\frac{1}{2}}\right|\right|_p \leq \left(\sum^k_i \left|\left|f_i\right|\right|^2_p\right)^{\frac{1}{2}},
\end{equation}
given the condition: $2\leq p < \infty, f_i \in L^p(E) (i=1,2,...,k)$.
Here is the snapshot .

My attempts are as follows:


*

*It suffices to consider $k=2$.

*Without loss of generality, assume all $f_i>0$.

*Put the $\frac{1}{2}$ on LHS out of the norm, and cancel it with the $\frac{1}{2}$ on RHS.

*Put the $2$ on RHS into the p-norm. It becomes:


\begin{equation}
||\sum^k_{i=1}f_i ||_{\frac{p}{2}} \leq \sum^k_i \left|\left| f_i^2 \right|\right|_{\frac{p}{2}},
\end{equation}
The condition $2\leq p < \infty$ reminds me the Clarkson inequalities, and in fact they appeared slightly before this problem. Yet the book didn't provide any hint. 
Then I'm stucked. Can anyone help me?

Thank you for the comments!
Though the book has rare typos, the original inequality may have square $2$ outside the LHS p-norm? That is,
\begin{equation}
\left|\left|\left(\sum^k_{i=1}|f_i|\right)^{\frac{1}{2}}\right|\right|_p^2 \leq \left(\sum^k_i \left|\left|f_i\right|\right|^2_p\right)^{\frac{1}{2}},
\end{equation}
such that the simplified one is
\begin{equation}
||\sum^k_{i=1}f_i ||_{\frac{p}{2}}^2 \leq \sum^k_i \left|\left| f_i^2 \right|\right|_{\frac{p}{2}},
\end{equation}
The reason is that, similar to Cauchy, Minkowski, Holder inequalities, both sides now have the same "order"/"power".
No matther which case is true, this kind of inequality is elegant. Hope we can achieve the right one.

@Calvin Khor I think your comment is correct. In this way, the question is to show:
\begin{equation}
\left|\left|\left(\sum^k_{i=1}|f_i|^2\right)^{\frac{1}{2}}\right|\right|_p \leq \left(\sum^k_i \left|\left|f_i\right|\right|^2_p\right)^{\frac{1}{2}},
\end{equation}
Then the proof is obvious!
 A: Following the comment of @Calvin Khor, and based on the difficulty of the problem it should be in this book, here I prove the following statement:
\begin{equation}
\left|\left|\left(\sum^k_{i=1}|f_i|^2\right)^{\frac{1}{2}}\right|\right|_p \leq \left(\sum^k_i \left|\left|f_i\right|\right|^2_p\right)^{\frac{1}{2}},
\end{equation}
given the condition that $2\leq p < \infty, f_i \in L^p(E) (i=1,2,...,k)$.
Proof: It suffices to show
\begin{equation}
\left|\left|\left(\sum^k_{i=1}|f_i|^2\right)\right|\right|^{\frac{1}{2}}_{\frac{p}{2}} \leq \left(\sum^k_i \left|\left|f_i\right|\right|^2_p\right)^{\frac{1}{2}},
\end{equation} 
\begin{equation}
\left|\left|\left(\sum^k_{i=1}|f_i|^2\right)\right|\right|_{\frac{p}{2}} \leq \sum^k_i \left|\left|f_i\right|\right|^2_p,
\end{equation} 
\begin{equation}
\left|\left|\left(\sum^k_{i=1}|f_i|^2\right)\right|\right|_{\frac{p}{2}} \leq \sum^k_i \left|\left| |f_i|^2\right|\right|_{\frac{p}{2}},
\end{equation} 
where the last inequality holds by Minkowski inequality. This completes the proof.
