This inequality has a long history. Starting from \frac{a+b}{2} \geq \sqrt{ab}, people gradually discovered that as long as a_1, a_2, \dots, a_n \geq 0, then it must hold that \frac{a_1+a_2+\dots+a_n}{n} \geq \sqrt[n]{a_1 a_2 \dots a_n}. For small values of n, it has already been proven, but the proof of the general form is a more recent development.
I encountered this inequality quite early (about three years ago, when I was in the sixth grade). Since then, I have been looking for a proof, but except for the case n=2, I remained unsuccessful. It was not until a math class three months ago, while daydreaming, that I suddenly figured it out (^_^). Initially, I proved the case for n=3, and then, with irresistible force, I proved that the inequality holds for any n.
In fact, the above inequality is equivalent to a_1^n + a_2^n + \dots + a_n^n \geq n a_1 a_2 \dots a_n. The purpose of this article is to prove this specific form. The method used is "Mathematical Induction," specifically strong induction: to prove the case for n=k, we assume it holds for n=k-1, n=k-2, \dots, n=1. Since we can prove the case for n=2 directly, the inequality is established step by step. The proof is described as follows (where letters like a, b, c represent non-negative numbers):
Before describing the general proof, let us look at a few examples.
(1) Prove: a^2 + b^2 \geq 2ab
Let us set aside the textbook proof. Suppose a \leq b and let b = a + x. Then we have a^2 + (a+x)^2 \geq 2a(a+x). Expanding this, we get 2a^2 + 2ax + x^2 \geq 2a^2 + 2ax, which is obviously true.
(2) Prove: a^3 + b^3 + c^3 \geq 3abc
Using the same method, let a \leq b \leq c, b = a + x, and c = a + y. Then we have a^3 + (a+x)^3 + (a+y)^3 \geq 3a(a+x)(a+y). Expanding both sides: a^3 + a^3 + 3a^2x + 3ax^2 + x^3 + a^3 + 3a^2y + 3ay^2 + y^3 \geq 3a^3 + 3a^2x + 3a^2y + 3axy This simplifies to: x^3 + y^3 + 3a(x^2 + y^2) \geq 3axy. Since x^2 + y^2 \geq xy is known, this expression is clearly true.
After seeing these, do you understand the key to the proof? The general case is as follows:
Elementary Mathematics Version
Assume that for values 1, 2, \dots, n-1, the inequality holds. We now prove: a_1^n + a_2^n + \dots + a_n^n \geq n a_1 a_2 \dots a_n.
Let a_{i+1} \geq a_i, a_1 = a, and a_{i+1} = a + x_i for i = 1, 2, \dots, n-1. The expression becomes: a^n + (a+x_1)^n + \dots + (a+x_{n-1})^n \geq na(a+x_1)\dots(a+x_{n-1}) Expanding using Pascal’s Triangle: \begin{aligned} & a^n + \\ & a^n + C_n^1 a^{n-1}x_1 + C_n^2 a^{n-2}x_1^2 + \dots + x_1^n + \\ & a^n + C_n^1 a^{n-1}x_2 + C_n^2 a^{n-2}x_2^2 + \dots + x_2^n + \\ & \dots + \\ & a^n + C_n^1 a^{n-1}x_{n-1} + C_n^2 a^{n-2}x_{n-1}^2 + \dots + x_{n-1}^n \geq \\ & na^n + na^{n-1}(x_1 + x_2 + \dots + x_{n-1}) + na^{n-2}(x_1 x_2 + x_1 x_3 + \dots + x_{n-2}x_{n-1}) + \\ & \dots + nax_1 x_2 \dots x_{n-1} \end{aligned}
Note: Here C_a^b refers to the number of combinations, which are the values in Pascal’s Triangle. It is defined as the number of ways to choose b distinct items from a set of a items, calculated as: C_a^b = \frac{a!}{b!(a-b)!}
Canceling and combining terms, we have: \begin{aligned} & \frac{C_n^2}{n} a^{n-2}(x_1^2 + \dots + x_{n-1}^2) + \frac{C_n^3}{n} a^{n-3}(x_1^3 + \dots + x_{n-1}^3) + \dots + \frac{x_1^n + x_2^n + \dots + x_{n-1}^n}{n} \\ & \geq a^{n-2}(x_1 x_2 + \dots + x_{n-2}x_{n-1}) + a^{n-3}(x_1 x_2 x_3 + \dots + x_{n-3}x_{n-2}x_{n-1}) + \dots \\ & + ax_1 x_2 \dots x_{n-1} \end{aligned} \tag*{\textcolor{green}{(A)}}
To prove the original inequality, we must prove that the above expression holds.
The number of terms in (x_1 x_2 + \dots + x_{n-2}x_{n-1}) is C_{n-1}^2, and the number of terms in (x_1 x_2 x_3 + \dots + x_{n-3}x_{n-2}x_{n-1}) is C_{n-1}^3.
Through cyclic summation, we know that when a_1^p + a_2^p + \dots + a_p^p \geq p a_1 a_2 \dots a_p holds, it must follow that \frac{C_{n-1}^p}{n-1}(x_1^p + \dots + x_{n-1}^p) \geq (x_1 x_2 \dots x_p + \dots + x_{n-p}x_{n-p+1} \dots x_{n-1}).
Furthermore, we have \frac{C_n^p}{n} \geq \frac{C_{n-1}^p}{n-1}. Therefore, in (A), each term on the left is no less than the corresponding term on the right, so inequality (A) holds. Thus, a_1^n + a_2^n + \dots + a_n^n \geq n a_1 a_2 \dots a_n is proven.
After reading this, does it seem complicated? Believe me, if you read it carefully, you will find it quite simple. Especially if used to prove cases for small n, this is absolutely the most perfect method!
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