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Algebra Difficulty 4.9 AIME Prove it Czech-Polish-Slovak Mathematical Match

Let nn be a given positive integer. Solve the system of equations
x1+x22+x33++xnn=n,x1+2x2+3x3++nxn=n(n+1)2 \begin{aligned} x_1 + x_2^2 + x_3^3 + \dots + x_n^n &= n, \\ x_1 + 2x_2 + 3x_3 + \dots + nx_n &= \frac{n(n+1)}{2} \end{aligned}
in the set of nonnegative real numbers x1,x2,,xnx_1, x_2, \dots, x_n.

Solution

Suppose x1,x2,,xnx_1, x_2, \dots, x_n satisfy the equations above. Then we have
0=x1+x22+x33++xnnn(x1+2x2+3x3++nxn12n(n+1))=(x222x2+21)+(x333x3+31)++(xnnnxn+n1). \begin{aligned} 0 &= x_1 + x_2^2 + x_3^3 + \dots + x_n^n - n - (x_1 + 2x_2 + 3x_3 + \dots + nx_n - \frac{1}{2}n(n+1)) \\ &= (x_2^2 - 2x_2 + 2 - 1) + (x_3^3 - 3x_3 + 3 - 1) + \dots + (x_n^n - nx_n + n - 1). \end{aligned}
However, the expressions in the brackets are nonnegative. Indeed, for k2k \ge 2 and x0x \ge 0 we have, by the AM-GM inequality,
xk+k1=xk+1+1++1kxkk=kx x^k + k - 1 = x^k + 1 + 1 + \dots + 1 \ge k \cdot \sqrt[k]{x^k} = kx
and the equality holds if and only if x=1x = 1. Therefore we have x2=x3==xn=1x_2 = x_3 = \dots = x_n = 1 and, by the first equation, x1=1x_1 = 1.

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