maximum when $x_i=x,x=n^{1/(n-1)}$ and $S=n^{n/(n-1)}.$

Rust wrote: maximum when $x_i=x,x=n^{1/(n-1)}$ and $S=n^{n/(n-1)}.$ Sorry, but this is twice wrong : 1) wrong because problem statement says "$x_i$" run through the positive integers" ... 2) wrong because even without this constraint, $n^{\frac n{n-1}}$ is far from the maximum value since $n^{\frac n{n-1}}<2n$ $\forall n>2$ and $\{x_i\}=\{1,1,1,...,1,2,n\}$ is a solution giving $\sum x_i=\prod x_i=2n$ In fact, without the "integer" constraint, there is obviously no maximum for both sides of this equation

You are right, I found minimal value. Let $S_i=\sum_{j=1}^i x_j, P_i=\prod_{j=1}^ix_j.$ Then $x_n=\frac{S_{n-1}}{P_{n-1}-1}.$ Because $x_n\ge 2$ we get $S_{n-1}\ge 2(P_{n-1}-1)\ge P_{n-1}$. $S_{n-1}\ge P_{n-1}\to S_{n-2}\ge x_{n-1}(P_{n-2}-1).$ While $x_i\ge 2\to S_{i-1}\ge P_{i-1}.$ If $x_1=2$, then $n=2,x_2=2$. All other solutions are $x_1=x_2=...=x_k=1, 2\le x_{k+1}\le x_{k+2}\le...\le x_n$. were $k=\prod_{i=k+1}^nx_i-\sum_{i=k+1}^nx_i, k\le n-2.$ In this way we find for given $S=S_n=P_n$ solution by factorization $S=(y_1+1)....(y_l+1),1\le y_1\le y_2\le ...\le y_l$ we find $k=S-(y_1+1)-(y_2+1)-...-(y_l+1), n=k+l=S-y_1-y_2---y_l$. Obviosly $S$ is not prime. If $l=2$, then $S=(a+1)(b+1),n=(a+1)(b+1)-a-b=ab+1\to S\le 2n$. If $l\ge 3$, then $S=(y_1+1)...(y_l+1),n=S-y_1-y_2...-y_l$ and $$S=n\frac{1}{1-x}, x=\frac{\sum_i y_i}{\prod_i (1+y_i)}\le frac 12 \to S\le 2n$. $S=2n$ only if $x_n=n,x_{n-1}=2, x_i=1,i<n-1.$

Could anyone solve it?

Juraev wrote: Could anyone solve it? Have you just read the previous posts ? We proved that there is no such maximum (looking for example at $\{x_i\}=\{1,1,1,...,1,2,n\}$

@pco: I think the problem is to determine for any fixed $n\ge2$ the corresponding maximal value $V(n)$ for this particular choice of $n$. Then for instance, $V(2)=4$.

Any solution on the maximum?

Suppose $x_1 = x_2 =... = x_k = 1 < 2 \le x_{k+1}\le ... \le x_n$. We have $M =x_1...x_n=x_{k+1} ... x_n$ $= (x_{k+1}- 1)x_{k+2}...x_n + x_{k+2} ... x_n$ $\geq (x_{k+1}- 1).2 + x_{k+2} ... x_n$ $\geq .................... .$ $\geq (x_{k+1}- 1).2+...+ (x_{n-2}- 1).2+x_{n-1}.x_n$ $\geq (x_{k+1}- 1).2+...+ (x_{n-2}- 1).2+(x_{n-1}- 1).2+(x_{n}- 1).2$ $= 2(x_{k+1} + ... + x_n- (n- k))$ $= 2(M- n).$ Therefore, $M \le 2n$. The inequality holds when $x_1=x_2=x_{n-2}=1$, $x_{n-1}=2$ and $x_n=n$