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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 $a_1, a_2, dots, a_n in BbbR$.

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 $a_1, a_2, cdots, a_n$,
$$fraca_1 + a_2 + cdots + a_nn ge sqrt[n]a_1 a_2 cdots a_n.$$
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 $n=1$, $b_1^1 = b_1 ge 1cdot b_1.$

(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,

$$
left{
beginarrayc
c_1(c_2^k + c_3^k + cdots + c_k+1^k) ge c_1(k c_2 c_3 cdots c_k+1) = k prod_j=1^k+1 c_j \
c_2(c_1^k + c_3^k + cdots + c_k+1^k) ge c_2(k c_1 c_3 cdots c_k+1) = k prod_j=1^k+1 c_j \
cdots \
c_k+1(c_1^k + c_2^k + cdots + c_k^k) ge c_k+1(k c_1 c_2 cdots c_k) = k
prod_j=1^k+1 c_j,\
endarray
right.
$$
i.e., for $1 le i le k+1$,
$$sum_j=1^k+1delta_ij^' c_j^k c_i ge k prod_j=1^k+1 c_j$$
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

$$sum_i=1^k+1 c_i^k+1 ge (k+1) prod_j=1^k+1 c_j.$$

There is the following reasoning for even $n$. $$nsum_k=1^na_k^n-sum_k=1^na_k^n-1sum_k=1^na_k=sum_1leq i<jleq nleft(a_i^n-a_i^n-1a_j-a_ia_j^n-1+a_j^nright)=$$
$$=sum_1leq i<jleq nleft(a_i^n-1-a_j^n-1right)left(a_i-a_jright)geq0.$$

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 $$(n,0,0,...,0)succ(n-1,1,0,...,0).$$