Let $X_1, X_2,\dots$ be independent, uniform$(0,1)$ random variables.
By the law of large numbers we have
$$\begin{eqnarray*}
{X_1+\cdots + X_n\over n}&\to& \mathbb{E}(X)={1\over 2}\\
{X_1^2+\cdots + X^2_n\over n}&\to& \mathbb{E}(X^2)={1\over 3}\\
\end{eqnarray*}
$$
in probability as $n\to\infty$. Therefore
$${X_1^2+\cdots +X^2_n\over X_1+\cdots +X_n}={X_1^2+\cdots +X^2_n\over n}\cdot{n\over X_1+\cdots +X_n}\to {2\over 3}$$ in probability as $n\to \infty$.
The ratio random variables ${X_1^2+\cdots +X^2_n\over X_1+\cdots +X_n}$ are bounded below by zero and above by one. This guarantees convergence of the expectations, as well.
So $$\mathbb{E}\left({X_1^2+\cdots +X^2_n\over X_1+\cdots +X_n}\right)\to{2\over 3}$$
which is the required result.
The probability approach makes things simpler. Let $X_1,X_2,\dots,X_n,\dots$ be independent random variables, which are uniform on $[0,1]$.
$X_1$ (or each other member in the family) has mean $\mu=\frac 12$ and variance $\sigma^2=\frac 1{12}$.
Let $Y_n$ be the mean $$Y_n=\frac 1n(X_1+X_2+\dots+X_n)\ .$$
Let $\mu$ be $1/2$ for short, and
$$
f(s) =
\frac 1{0!}f(\mu) +
\frac 1{1!}f'(\mu)(s-\mu) +
\frac 1{2!}f''(\mu)(s-\mu)^2 +
g(s)\ ,
$$
where $g(s)$ has an estimation of the shape $|g(s)|\le M\;|s-\mu|^3$ for some multiplicative constant $M$. We denote by $\Bbb E$ the expectation on the probability space where all variables $(X_n)$ live in.
Then let us denote by $I_n$ the integral:
$$
\begin{aligned}
I_n&=
\int_0^1\int_0^1\cdots\int_0^1 n
\left[\
f\left(\frac{x_1+\cdots+x_n}{n}\right)-f\left(\frac 12\right)
\ \right]
\,dx_1\,dx_2\cdots\,dx_n
\\
&=\Bbb E[\ n(f(Y_n)-f(\mu))\ ]
\\
&=\Bbb E\left[\ n\left(
\frac 1{1!}f'(\mu)(Y_n-\mu) +
\frac 1{2!}f''(\mu)(Y_n-\mu)^2 +
g(Y_n)
\right)\ \right]
\\
&=
\frac 1{1!}f'(\mu)\cdot n\;\underbrace{\Bbb E[\ (Y_n-\mu)\ ]}_{=0} +
\frac 1{2!}f''(\mu)\cdot n\;\underbrace{\Bbb E[\ (Y_n-\mu)^2\ ]}_{\operatorname{Var}[Y_n]=\sigma^2/n} +
\underbrace{n\;\Bbb E[\ g(Y_n)\ ]}_{\to 0}
\\
&\to\frac 1{2!}f''(\mu)\cdot\sigma^2=\frac 12f''(\mu)\cdot\frac 1{12}=\frac 1{24}f''(\mu)\ .
\end{aligned}
$$
Here, we have used the estimation for $g$, and the limit
$$
n\Bbb E[\ |Y_n-\mu|^3\ ]=
n\|\ (Y_n-\mu)^3\ \|_{L^1}
\le
n
\cdot
\|\ (Y_n-\mu)^1\ \|_{L^2}
\cdot
\|\ (Y_n-\mu)^2\ \|_{L^2}
\to 0\ .
$$
So the result is
$$
\lim_{n\to\infty}I_n=\frac 1{12}f''\left(\frac 12\right)\ .
$$
Best Answer
Using the AG-GM :
If $x_1,x_2,\cdots,x_n$ are positive real numbers, then
$$n\sqrt[n]{x_1^2x_2^2\cdots x_n^2}\leq x_1^2+x_2^2+ \cdots x_n^2$$
You can write: $$ \int_1^\infty \cdots \int_1^\infty \frac{dx_1 \cdots dx_n}{(x_1^{2}+\cdots + x_n^{2})^a} \leq \frac{1}{n} \int_{1}^\infty \cdots \int_{1}^\infty \frac{dx_1 \cdots dx_n}{x_1^{2a/n}\cdots x_n^{2a/n}} $$ But: $$ \frac{1}{n} \int_{1}^\infty \cdots \int_{1}^\infty \frac{dx_1 \cdots dx_n}{x_1^{2a/n}\cdots x_n^{2a/n}} = \frac{1}{n} \int_{1}^\infty \frac{dx_1}{x_1^{2a/n}} \cdots \int_{1}^\infty \frac{dx_n}{x_n^{2a/n}} $$ And, since $\frac{2a}{n}> 1$,for each $i \in [1,\infty)$: $$ \int_{1}^\infty \frac{dx_i}{x_i^{2a/n}} = \left[ \frac{x_i^{1-2a/n}}{1-2a/n} \right]_1^\infty = \frac{1}{1-2a/n} $$