Prove the inequality $\left(\frac n{n-1}\right)^{n-1}(\frac1n\sum_{i=1}^na^2_i)+(\frac1n\sum_{i=1}^nb_i)^2\ge\prod_{i=1}^n(a^2_i+b^2_i)^{\frac1n}$ Prove that for $n\ge 2$, $n\in \mathbb{N}$, $a_{i}, b_{i}\ge 0$
$$\left(\dfrac{n}{n-1}\right)^{n-1}\left(\dfrac{1}{n}\displaystyle\sum_{i=1}^{n}a^2_{i}\right)+\left(\dfrac{1}{n}\displaystyle\sum_{i=1}^{n}b_{i}\right)^2\ge\displaystyle\prod_{i=1}^{n}(a^2_{i}+b^2_{i})^{\frac{1}{n}}$$
When $n=2$, I can easily show that
$$a^2_{1}+a^2_{2}+\dfrac{1}{4}(b_{1}+b_{2})^2\ge\sqrt{(a^2_{1}+b^2_{1})(a^2_{2}+b^2_{2})}$$
I want to use Induction to prove this. Any suggestions on how to solve it?
 A: Fix $a_2,\cdots,a_n, b_2,\cdots, b_n$. Consider $b_1' = t$, $a_1'^2 = a_1^2 + b_1^2 - t^2$, for $t \in [0,\sqrt{a_1^2+b_1^2}]$. With this pair $(a_1',b_1')$ replacing $(a_1,b_1)$, one sees that right hand side is unchanged, and left hand side is a concave quadratic function in $t$. This means that LHS attains its minimum when $t$ hits the boundary, i.e. $(a_1,b_1)$ being replaced by $(\sqrt{a_1^2+b_1^2}, 0)$, or $(0, \sqrt{a_1^2+b_1^2})$.
This reduces the problem to the following one. Let $x_i^2 = a_i^2 + b_i^2$, $x_i \ge 0$. For each index $i$, $(a_i,b_i) = (x_i,0)$ or $(0,x_i)$. Rearrange the indices so that the first $k$ indices satisfy $(a_i,b_i) = (x_i,0)$, and the last $(n-k)$ satisfies $(a_i,b_i) = (0,x_i)$. We then need to show 

$$\left(\dfrac{n}{n-1}\right)^{n-1}\left(\dfrac{1}{n}\displaystyle\sum_{i=1}^{k}x_i^2\right)+\left(\dfrac{1}{n}\displaystyle\sum_{i=k+1}^{n}x_i\right)^2\ge\displaystyle\prod_{i=1}^{n}x_i^{\frac{2}{n}}$$
for any $x_i \ge 0$.

Apply AM-GM for the two terms on the left separately, we get LHS is at least 
$$\frac{k}{n} \left(\dfrac{n}{n-1}\right)^{n-1} (x_1\cdots x_k)^{1/k} + \frac{(n-k)^2}{n^2} (x_{k+1}\cdots x_n)^{1/(n-k)}$$
The inequality is clearly true when $k = 0$ or $n$. Otherwise apply AMGM again, with suitable weights, we get that this is at least
$$\left(\dfrac{n}{n-1}\right)^{k(n-1)/n}\left(\frac{n-k}{n}\right)^{(n-k)/n}(x_1\cdots x_n)^{2/n}$$
So it suffices to show that 
$$\left(\dfrac{n}{n-1}\right)^{n-1}\left(\frac{n-k}{n}\right)^{(n-k)/k} \ge 1$$
Equivalently,
$$\left(\dfrac{n}{n-1}\right)^{n-1} \ge \left(\frac{n}{n-k}\right)^{(n-k)/k} = \left(\frac{1}{1 - k/n}\right)^{n/k - 1}$$
Hence consider the function $f(u) = (1-u)^{u-1}$ for $u \in (0,1)$. It suffices to show that $f(u)$ is increasing in this range, which is true from the plot at wolframalpha.
