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Algebra Difficulty 7.1 National olympiad, round 2 Prove it

4. If x1,x2,⋯ ,xnx_{1}, x_{2}, \cdots, x_{n} are all positive, then x1n+1+x2n+1+⋯+xnn+1⩾x1x2⋯xn(x1+x_{1}^{n+1}+x_{2}^{n+1}+\cdots+x_{n}^{n+1} \geqslant x_{1} x_{2} \cdots x_{n}\left(x_{1}+\right. x2+⋯+xn)\left.x_{2}+\cdots+x_{n}\right). (College Mathematics Journal, Vol. 25, No. 4, 1994)

Solution

4. By the generalization of the Cauchy-Schwarz inequality, we have
(x1n+1+x2n+1+⋯+xnn+1)(x1n+1+x2n+1+⋯+\left(x_{1}^{n+1}+x_{2}^{n+1}+\cdots+x_{n}^{n+1}\right)\left(x_{1}^{n+1}+x_{2}^{n+1}+\cdots+\right.
xnn+1)(x2n+1+x3n+1+⋯+xnn+1+x1n+1)⋯(xnn+1+x1n+1+x2n+1+⋯+xn−1n+1)⩾(x12x2⋯xn+x1x22⋯xn+⋯+x1x2⋯xn2)n+1=[x1x2⋯xn(x1+x2+⋯+xn)]n+1\begin{array}{l} \left.x_{n}^{n+1}\right)\left(x_{2}^{n+1}+x_{3}^{n+1}+\cdots+x_{n}^{n+1}+x_{1}^{n+1}\right) \cdots\left(x_{n}^{n+1}+x_{1}^{n+1}+x_{2}^{n+1}+\cdots+x_{n-1}^{n+1}\right) \geqslant \\ \left(x_{1}^{2} x_{2} \cdots x_{n}+x_{1} x_{2}^{2} \cdots x_{n}+\cdots+x_{1} x_{2} \cdots x_{n}^{2}\right)^{n+1}=\left[x_{1} x_{2} \cdots x_{n}\left(x_{1}+x_{2}+\cdots+x_{n}\right)\right]^{n+1} \end{array}

Taking the (n+1)(n+1)-th root on both sides, we get
x1n+1+x2n+1+⋯+xnn+1⩾x1x2⋯xn(x1+x2+⋯+xn)x_{1}^{n+1}+x_{2}^{n+1}+\cdots+x_{n}^{n+1} \geqslant x_{1} x_{2} \cdots x_{n}\left(x_{1}+x_{2}+\cdots+x_{n}\right)

Equality holds if and only if x1=x2=⋯=xnx_{1}=x_{2}=\cdots=x_{n}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.