Collection of Maths Problems

AG inequality

Task number: 2799

• Variant 1

  For \(n=2\).

• Resolution

  \( \displaystyle \sqrt{x_1x_2}\le \frac{x_1+x_2}2 \iff 2\sqrt{x_1x_2}\le x_1+x_2 \iff 4x_1x_2 \le x_1^2 + 2x_1x_2+x_2^2 \iff \\ x_1^2 - 2x_1x_2+x_2^2\ge 0 \iff (x_1-x_2)^2 \ge 0 \)

  Non-negativity is necessary for two reasons: the expression \(x_1x_2\) below the root must be non-negative and also, for the second step to be valid, \(x_1+x_2\) must also be non-negative.

• Variant 2

  When \(n\) is a power of two.

• Resolution

  We assume that the statement holds for \(n\) and \(n=2\), and prove it for \(2n\).

  \( \displaystyle \sqrt[2n]{x_1\cdots x_{2n}} =\sqrt{\sqrt[n]{x_1\cdots x_{n}}\sqrt[n]{x_{n+1}\cdots x_{2n}}} \le \frac12\left(\sqrt[n]{x_1\cdots x_{n}}+\sqrt[n]{x_{n+1}\cdots x_{2n}}\right)\le \\ \le \frac12\left(\frac{x_1+…+x_n}n+\frac{x_{n+1}+…+x_{2n}}n\right) =\frac{x_1+…+x_{2n}}{2n}\)

  In the first inequality we used the assumption for \(n=2\), in the second for \(n\).

• Variant 3

  For any \(n\in \mathbb N\).

• Hint

  Show that if the statement holds for \(n+1\) then it is also true for \(n\).
  Choose \(x_{n+1}\) so that you can easily reduce the given expression on the left side.

• Resolution

  First we transform

  \( \displaystyle \sqrt[n+1]{x_1x_2\cdots x_{n+1}} \le \frac{x_1+x_2+…+x_{n+1}}{n+1} \iff (n+1)^{n+1}{x_1x_2\cdots x_{n+1}} \le (x_1+x_2+…+x_{n+1})^{n+1} \)

  We choose \(x_{n+1}=\frac1n(x_1+…+x_n)\), which gives us

  \( \displaystyle \frac{(n+1)^{n+1}}nx_1x_2\cdots x_n(x_1+x_2+…+x_n) \le \left(\left(1+\frac1n\right)(x_1+x_2+…+x_n)\right)^{n+1} \iff \\ \frac{(n+1)^{n+1}}n x_1x_2\cdots x_n \le \left(\frac{n+1}n\right)^{n+1}(x_1+x_2+…+x_n)^n \iff\\ n^{n}{x_1x_2\cdots x_n} \le (x_1+x_2+…+x_n)^{n}\iff \sqrt[n]{x_1x_2\cdots x_{n}} \le \frac{x_1+x_2+…+x_{n}}{n} \)