For all postive real numbers $a_{1},a_{2},\cdots,a_{n}$,the following inequality holds
\[ \sum_{k=1}^{n}{a^{\frac{k}{k+1}}_{k}}\leq \sum_{k=1}^{n}{a_{k}}+\sqrt{\frac{2(\pi^2-3)}{9}\sum_{k=1}^{n}{a_{k}}} \]
\emph{Proof}:We begin with a preliminary result.\\
\textbf{Lemma}.If
\[ m_{k}=\left\{
     \begin{array}{ll}
       1, & \hbox{if \ $k=1$;} \\
       2\sqrt{\frac{(k-1)^{k-1}}{(k+1)^{k+1}}}, & \hbox{if \ $k>1$ .}
     \end{array}
   \right.
\]
then
\[ a_{k}^{\frac{k}{k+1}}\leq a_{k}+m_{k}\sqrt{a_{k}} \]
Proof,If $k=1$,this is trivially true.Otherwise,if $k>1$,the inequality can be written as
\[ a_{k}^{\frac{k}{k+1}}\left(1-a^{\frac{1}{k+1}}_{k} \right)\leq m_{k}\sqrt{a_{k}} \]
For the nontrivial case $0<a_{k}<1$,this inequality is equivalent to
\[ a^{\frac{k-1}{k+1}}_{k}\left(1-a^{\frac{1}{k+1}}_{k} \right)^{2}\leq m^{2}_{k} \]
To prove it,we use the AM-GM Inequality,as follows\\
\begin{align*}
a^{\frac{k-1}{k+1}}_{k}\left(1-a^{\frac{1}{k+1}}_{k}\right)^{2}
&=4(k-1)^{k-1}\left(\frac{a^{\frac{1}{k+1}}_{k}}{k-1}\right)^{k-1}\left(\frac{1-a^{\frac{1}{k+1}}_{k}}{2}\right)^{2}\\
&\leq4(k-1)^{k-1}\left[\frac{(k-1)\left(\frac{a^{\frac{1}{k+1}}_{k}}{k-1}\right)+2\left(\frac{1-a^{\frac{1}{k+1}}_{k}}{2}\right)}{k+1}\right]^{k+1}\\
&=\frac{4(k-1)^{k-1}}{(k+1)^{k+1}}\\
&=m^{2}_{k}
\end{align*}
and so the lemma is proved.\\
Returning to our problem,we see that it suffices to show that
\[ \sum_{k=1}^{n}{m_{k}\sqrt{a_{k}}}\leq \sqrt{\frac{2(\pi^2-3)}{9}\sum_{k=1}^{n}{a_{k}}}\]
But the \emph{Cauchy-Schwarz} Inequality gives us that
\[ \sum_{k=1}^{n}{m_{k}\sqrt{a_{k}}}\leq \sqrt{\left(\sum_{k=1}^{n}{m^{2}_{k}}\right)\left(\sum_{k=1}^{n}a_{k}\right)}\]
So,it remains to prove that
\[ \sum_{k=1}^{n}{m^{2}_{k}}\leq \frac{2(\pi^2-3)}{9}\]
or
\[ \frac{1}{4}+\sum_{k=2}^{n}{\frac{(k-1)^{k-1}}{(k+1)^{k+1}}}\leq \frac{\pi^2}{18}-\frac{1}{6}\]
Now,because
\[ \frac{(k+1)^{k+1}}{(k-1)^{k-1}}=(k+1)^{2}\left(\frac{k+1}{k-1}\right)^{k-1}=(k+1)^2\left(1+\frac{2}{k-1}\right)^{k-1}\]
\emph{Bernoulli's} Inequality yields
\[ \frac{(k+1)^{k+1}}{(k-1)^{k-1}}\geq 3(k+1)^{2} \]
and thus
\[ \frac{1}{4}+\sum_{k=2}^{n}{\frac{(k-1)^{k-1}}{(k+1)^{k+1}}}\leq \frac{1}{4}+\frac{1}{3}\sum_{k=2}^{n}{\frac{1}{(k+1)^{2}}}=\frac{1}{3}\sum_{k=1}^{n+1}{\frac{1}{k^2}}-\frac{1}{6}\]
On the other hand
\[ \lim_{n\rightarrow \infty}{\left(1+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\right)}=\zeta{(2)}=\frac{\pi^2}{6}\]
and therefore it follows that
\[ \frac{1}{4}+\sum_{k=2}^{n}{\frac{(k-1)^{k-1}}{(k+1)^{k+1}}}\leq \frac{\pi^2}{18}-\frac{1}{6}\]
which finishes our proof.Notice also that equality does not occur.$\square$