# Mathematical Induction Inequality Proof with Two Initials

Usually, mathematical induction inequality proof requires one initial value, but in some cases, two initials are required, such as in the Fibonacci sequence. In this case, it is necessary to show two initials are working as the first step of the mathematical induction inequality proof, and two assumptions are to be placed for the third step.
The following worked example shows how to handle the mathematical induction inequality proof with two initials.

## Worked Example

A sequence $S_n$ is defined by $S_1 = 1, S_2 = 2$ and for $n \gt 2, S_n = S_{n-1} + (n-1) S_{n-2}$.

(a)   Prove $\sqrt{x} + x \ge \sqrt{x(1 + x)}$ for all real numbers $x \ge 0$.

\begin{aligned} \require{AMSsymbols} \require{color} (\sqrt{x} + x)^2 &= x + 2x\sqrt{x} + x^2 \\ (\sqrt{x} + x)^2 &\ge x + x^2 &\color{green} {2x\sqrt{x} \ge 0} \\ \sqrt{x} + x &\ge \sqrt{x+x^2} \\ \therefore \sqrt{x} + x &\ge \sqrt{x(1+x)} \end{aligned}

(b)   Hence, prove $S_n \ge \sqrt{n!}$ for all integers $n \ge 1$ using the result of (a).

Step 1: Show it is true for $S_1$ and $S_2$.
For $n = 1, S_1 = 1$ and $\sqrt{1!} = 1$.
For $n = 2, S_2 = 2$ and $\sqrt{2!} = 1.4142 \ldots$.
Therefore $S_n \ge \sqrt{n!}$ is true for $n = 1$ and $n = 2$.
Step 2: Assume that it is true for $n = k$ and $n = k-1$.
Note that two assumptions are required if there are two initial values.
They are, $S_{k} \ge \sqrt{k!}$ and $S_{k-1} \ge \sqrt{(k – 1)!}$.
Step 3: Show it is true for $n = k + 1$.
That is, $S_{k+1} \ge \sqrt{(k + 1)!}$.
\begin{aligned} \require{AMSsymbols} \require{color} S_{k+1} &= S_{k} + k S_{k-1} &\color{green} {\text{given}} \\ &\ge \sqrt{k!} + k \sqrt{(k-1)!} &\color{green} {\text{Assumed at Step 2}} \\ &= \sqrt{k(k-1)!} + k \sqrt{(k-1)!} &\color{green} {k! = k(k-1)!} \\ &= \sqrt{(k-1)!}(\sqrt{k} + k) &\color{green} {\text{factorise by } \sqrt{(k-1)!}} \\ &\ge \sqrt{(k-1)!} \sqrt{k(k + 1)} &\color{green} {\text{proved by part (a)}} \\ &= \sqrt{(k + 1)!} \\ \therefore S_{k+1} &\ge \sqrt{(k + 1)!} \end{aligned}
Therefore $S_n \ge \sqrt{n!}$ is true for all integers $n \ge 1$.