# Proving AM-GM via $n \cdot (a_1^n + a_2^n + \dots + a_n^n) \ge (a_1^{n-1} + a_2^{n-1} + \dots + a_n^{n-1}) \cdot (a_1 + a_2 + \dots + a_n)$

I want to prove the arithmetic–geometric mean inequality. To prove that, I need the following inequality:

Suppose that $n$ is an integer which is greater than or equal to $1$ and $a_1, a_2, \dots, a_n \in \Bbb{R}$.

Then, $$n \cdot (a_1^n + a_2^n + \dots + a_n^n) \ge (a_1^{n-1} + a_2^{n-1} + \dots + a_n^{n-1}) \cdot (a_1 + a_2 + \dots + a_n).\ \cdots (1)$$

For any n positive real numbers $a_1, a_2, \cdots, a_n$, $$\frac{a_1 + a_2 + \cdots + a_n}{n} \ge \sqrt[n]{a_1 a_2 \cdots a_n}.$$ This inequality is obviously equivalent to the next inequality:

For any n positive real numbers $b_1, b_2, \cdots, b_n$, $$b_1^{n} + b_2^{n} + \cdots b_n^{n} \ge n (b_1 b_2 \cdots b_n)$$

We prove the second inequality by induction.

(i) When $n=1$, $b_1^{1} = b_1 \ge 1\cdot b_1.$

(ii)When $n=k$, suppose that the above inequality holds.

Let $c_1, c_2, \cdots, c_{k+1}$ be $k+1$ positive real numbers. Then, by the assumption above,

$$\left\{ \begin{array}{c} c_1(c_2^{k} + c_3^{k} + \cdots + c_{k+1}^k) \ge c_1(k c_2 c_3 \cdots c_{k+1}) = k \prod_{j=1}^{k+1} c_j \\ c_2(c_1^{k} + c_3^{k} + \cdots + c_{k+1}^k) \ge c_2(k c_1 c_3 \cdots c_{k+1}) = k \prod_{j=1}^{k+1} c_j \\ \cdots \\ c_{k+1}(c_1^{k} + c_2^{k} + \cdots + c_{k}^k) \ge c_{k+1}(k c_1 c_2 \cdots c_{k}) = k \prod_{j=1}^{k+1} c_j,\\ \end{array} \right.$$ i.e., for $1 \le i \le k+1$, $$\sum_{j=1}^{k+1}\delta_{ij}^{'} c_j^k c_i \ge k \prod_{j=1}^{k+1} c_j$$ where, $\delta_{ij}^{'} = 1 - \delta_{ij}$, $\delta_{ij}$ is Kronecker's delta.

So,

$$\sum_{i=1}^{k+1} \sum_{j=1}^{k+1} \delta_{ij}^{'} c_j^k c_i \ge \sum_{i=1}^{k+1} k \prod_{j=1}^{k+1} c_j = k(k+1) \prod_{j=1}^{k+1} c_j.$$

$$\sum_{i=1}^{k+1} \sum_{j=1}^{k+1} \delta_{ij}^{'} c_j^k c_i = \sum_{i=1}^{k+1} \sum_{j=1}^{k+1} c_j^k c_i - \sum_{i=1}^{k+1} c_i^{k+1} = \sum_{j=1}^{k+1} c_j^k \cdot \sum_{i=1}^{k+1} c_i - \sum_{i=1}^{k+1} c_i^{k+1}.$$

If (1) holds, then

$$(k+1) \sum_{i=1}^{k+1} c_i^{k+1} \ge \sum_{j=1}^{k+1} c_j^k \cdot \sum_{i=1}^{k+1} c_i.$$

So,

$$k \sum_{i=1}^{k+1} c_i^{k+1} \ge \sum_{j=1}^{k+1} c_j^k \cdot \sum_{i=1}^{k+1} c_i - \sum_{i=1}^{k+1} c_i^{k+1} \ge k(k+1) \prod_{j=1}^{k+1} c_j.$$

When we divide both sides of the above inequality by k, we get

$$\sum_{i=1}^{k+1} c_i^{k+1} \ge (k+1) \prod_{j=1}^{k+1} c_j.$$

• Well well, this is Chebyshev's inequality given that a1 > a2 > ... more in that link. Apr 25, 2014 at 11:15
• I didn't know Chebyshev's inequality. Thank you very much for your answer. Apr 25, 2014 at 21:31

There is the following reasoning for even $$n$$. $$n\sum_{k=1}^na_k^n-\sum_{k=1}^na_k^{n-1}\sum_{k=1}^na_k=\sum_{1\leq i $$=\sum_{1\leq i
For odd $$n$$ it's wrong.
For example, take $$n=3$$.
We get $$3(a^3+b^3+c^3)\geq(a^2+b^2+c^2)(a+b+c),$$ which is wrong for $$b=c=0$$ and $$a=-1.$$
Also, we can say that it's true for non-negative variables by Muirhead because $$(n,0,0,...,0)\succ(n-1,1,0,...,0).$$
• What if we restrict to nonnegative $a_k$? It should work for all $n$ then. Mar 27, 2019 at 19:22
• @Wojowu For non-negative variables it's true for all $n\geq1$, of course . The proof is the same. Also, we can use Chebyshov here. Mar 27, 2019 at 19:25