
5. Let $n>k>1$ be positive integers. Determine all positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$ which satisfy

$$
\sum_{i=1}^{n} \sqrt{\frac{k a_{i}^{k}}{(k-1) a_{i}^{k}+1}}=\sum_{i=1}^{n} a_{i}=n
$$


## Solution 1

By A.M-G.M inequality we have

$$
\frac{(k-1) a_{i}^{k}+1}{k} \geq\left(a_{i}^{k(k-1)}\right)^{1 / k}=a_{i}^{k-1}
$$

which implies $\sqrt{\frac{k a_{i}^{k}}{(k-1) a_{i}^{k}+1}} \leq \sqrt{a_{i}}$. Hence

$$
\sum_{i=1}^{n} \sqrt{a_{i}} \geq \sum_{i=1}^{n} \sqrt{\frac{k a_{i}^{k}}{(k-1) a_{i}^{k}+1}}=\sum_{i=1}^{n} a_{i}=n
$$

But by Cauchy-Schwarz inequality we have

$$
\sum_{i=1}^{n} \sqrt{a_{i}} \leq \sqrt{n\left(\sum_{i=1}^{n} a_{i}\right)}=n
$$

Therefore

$$
n \geq \sum_{i=1}^{n} \sqrt{a_{i}} \geq \sum_{i=1}^{n} \sqrt{\frac{k a_{i}^{k}}{(k-1) a_{i}^{k}+1}}=\sum_{i=1}^{n} a_{i}=n
$$

and hence equality holds everywhere which implies $a_{i}=1$ for $i=1,2, \ldots, n$.

