Base Case, we take two positive real numbers

We have proved it for any two positive real numbers

We take 2 sets of positive real numbers
and
such that
and
Adding these two inequalities we get

Let be positive real numbers respectively

Therefore

Therefore using induction we can prove the AM GM inequality for since we have proved base case of

We have proven that the AM GM inequality holds true for numbers. If we can prove that the inequality being true for numbers implies that it is true for numbers, we can fill in the gaps between exponents of two and show that it holds true for all numbers.

I worked on this backwards on paper and that’s how I will be proving it here by using bi-implicative statements.

We need to prove that for any real numbers

If we multiply both sides of an inequality by a positive number they inequality holds and visa versa.
We multiply it by the positive number where

We observe that (expansion has been left as exercise for TA Hint: Multiple numerator and denominator by n)

Since is a positive real number, we can take a set of numbers where is . We know that the AM GM inequality holds for it since and therefore the last statement is true. Since these statements are bi-implicative, it must also hold for the first numbers of this set or all numbers in this set other than confirming our initial statement:

\sqrt[n]{\prod a{i}} \leq \frac{\sum a{i}}{n}

\sqrt[n]{\prod{i=1}^{n}i} \leq \frac{\sum{i=1}^n i}{n}

\sqrt[n]{n!} \leq \frac{\frac{n(n+1)}{2}}{n} \implies \sqrt[n]{n!} \leq \frac{n+1}{2} \implies {n!} \leq \left( \frac{n+1}{2} \right)^n
$$