By AM GM and CS we have $$\sum_{i=1}^n \sqrt{\frac{k a_i^k}{(k-1) a_i^k+1}} \leq \sum_{i=1}^n \sqrt{\frac{k a_i^k}{k a_i^{k-1}}} \leq \sum_{i=1}^n \sqrt{a_i} \leq \sqrt{n(\sum_{i=1}^n a_i)} = n$$Therefore equality holds when $a_i=1$ for all $i$ $\blacksquare$

Yes 1 confirm

Deleted..

User2411 wrote: It is direct by Jensen' s inequality The function is not convex or concave for k > 2

Sg48 wrote: User2411 wrote: It is direct by Jensen' s inequality The function is not convex or concave for k > 2 yeah , it is concave for k=1 and 2 only

Claim: $ \sqrt \frac{kx^k}{(k-1)x^k + 1} \leq \frac{x+1}{2} $ for $ x \geq 0 $ and $k \geq 2 $ (equality holds at $x=1$ )

Hence $$\sum_{i=1}^n \sqrt{\frac{k a_i^k}{(k-1) a_i^k+1}} \leq \sum_{i=1}^n \frac{1+a_i}{2}=n $$Therefore equality holds when $a_i=1$ for all $1\leq i \leq n$

im kinda new to olympiad stuff, so here's an attempt... do correct me if im wrong we claim that $a_i=1$ is the only solution for $1 \leq{i} \leq{n}$. on substituing it we see that it works. now we show that its the only solution let $a_i>1$ $<=> a_i^k>1$ $<=> k\cdot a_i^k = (k-1)a_i^k+a_i^k > (k-1)a_i^k + 1 $ ( note that we can only do this because k>1) $<=>\frac{( k\cdot a_i^k)} {((k-1)a_i^k + 1)} > 1 $ ( in the previous line rhs is greater than 1) $<=>\sqrt{\frac{( k\cdot a_i^k)} {((k-1)a_i^k + 1)} } >1$ $<=>\sum_{k=1}^n\sqrt{\frac{( k\cdot a_i^k)} {((k-1)a_i^k + 1)} } >n$ which does not satisfy our requirement now we consider $a_i<1$ $<=> a_i^k<1$ $<=>(k-1)a_i^k < k-1$ $<=> (k-1)a_i^k + 1 < k$ $<=>\frac{ 1}{((k-1)a_i^k +1)}>\frac{1}{k}$ $<=> \frac{k}{((k-1)a_i^k +1)}>1>a_i^k$ $<=> \frac{k\cdot a_i^k}{((k-1)a_i^k +1)}> a_i^{2k}$ ( this is true because of the previous line) $<=>\sqrt{ \frac{k\cdot a_i^k}{((k-1)a_i^k +1)}}>\sqrt{ a_i^{2k}} = a_i^k$ $<=>\sum_{k=1}^n\sqrt{ \frac{k\cdot a_i^k}{((k-1)a_i^k +1)}}>\sum_{k=1}^n\sqrt{ a_i^{2k}} = \sum_{k=1}^na_i^k$ which again doesnt satisfy our requirement. hence proved.

Claim 1: $\sum_{i=1}^n \sqrt{\frac{k a_i^k}{(k-1) a_i^k+1}} \leq \sum_{i=1}^n \sqrt{a_i}$ From AM-GM inequality, we get: $$\frac{(k-1)a_i^k+1}{k}\geq a_i^{k-1} \implies \sqrt{(k-1)a_i^k+1}\geq \sqrt{ka_i^{k-1}}$$$$\implies \sqrt{\frac{k a_i^k}{(k-1) a_i^k+1}} \geq \sqrt{\frac{ka_i^k}{ka_i^{k-1}}}$$$$\implies \sqrt{\frac{k a_i^k}{(k-1) a_i^k+1}} \geq \sqrt{a_i}$$$$\implies\sum_{i=1}^n \sqrt{\frac{k a_i^k}{(k-1) a_i^k+1}} \leq \sum_{i=1}^n \sqrt{a_i}$$ Claim 2: $\sum_{i=1}^n \sqrt{a_i} \leq n$ We apply the QM-AM inequality on the $\sqrt{a_i}$'s, which gives: $$\sqrt\frac{a_1+a_2+\cdots+a_n}{n}\geq\frac{\sqrt{a_1}+\sqrt{a_2}+\cdots+\sqrt{a_n}}{n}$$$$\implies \sum_{i=1}^n \sqrt{a_i}\leq n\quad\quad\quad{\text{as }a_1+a_2+\cdots+a_n=n}$$ Therefore, combining the two, we get: $$\implies\sum_{i=1}^n \sqrt{\frac{k a_i^k}{(k-1) a_i^k+1}} \leq \sum_{i=1}^n \sqrt{a_i}\leq n$$ Therefore, for the equation to be valid, the equality case must hold, which is attained if and only if $a_i=1$ for all $i$, and we're done!

I have discussed this question in my RMO 2023 video on my channel "little fermat". Here is the Video