By considering the first three terms of the binomial expansion of \(\left[ 1+\left(\dfrac{1}{\sqrt{n}}\right) \right]^n\), where \(n\) is an integer greater than \(1\), prove that \[\left[1+\left(\frac{1}{\sqrt{n}}\right)\right]^{n-2} \geq \tfrac{1}{2}n.\]

Since \(n \geq 2\), we know that we have at least three terms in the binomial expansion (which in general has \(n+1\) terms).

Furthermore, we know every term is positive, since \(1\) and \(1/\sqrt{n}\) both are, which means we can truncate the sum to obtain an inequality. Hence \[\begin{align*} \left[ 1+\frac{1}{\sqrt{n}} \right]^n &= \sum_{k=1}^n {n \choose k}\left(\frac{1}{\sqrt{n}}\right)^k \\ &= 1+n\frac{1}{\sqrt{n}}+\frac{n(n-1)}{2}\left(\frac{1}{\sqrt{n}}\right)^2+\cdots \\ &\geq 1+n\frac{1}{\sqrt{n}}+\frac{n(n-1)}{2}\left(\frac{1}{\sqrt{n}}\right)^2\\ &= 1+\sqrt{n}+\frac{n-1}{2}. \end{align*}\]This looks promising, but the inequality we are trying to prove has \(n-2\) instead of \(n\) in the exponent on the left hand side.

To fix this, there are (at least) two approaches we can take.

#### Approach 1: Divide through by \((1+1/\sqrt{n})^2\)

The most direct approach is to simply divide both sides of our inequality by \((1+1/\sqrt{n})^2\).

After this, the left hand side is what we want, but the right hand side is a bit of a mess: \[ \left[ 1+\frac{1}{\sqrt{n}} \right]^{n-2} \geq \frac{1+\sqrt{n}+\frac{n-1}{2}}{\left(1+\frac{1}{\sqrt{n}}\right)^2}. \]

To clean it up, we simplify the numerator and denominator of the right hand side. The numerator can be written over a common denominator to get

\[1+\sqrt{n}+\frac{n-1}{2}=\frac{1+2\sqrt{n}+n}{2}= \frac{1}{2}(1+\sqrt{n})^2.\]

(For the final step here, we noticed that \(1+2\sqrt{n}+n\) is the square of \(1+\sqrt{n}\).)

Similarly, the denominator can be expanded to get \[\left(1+\frac{1}{\sqrt{n}}\right)^2= \left(\frac{\sqrt{n}+1}{\sqrt{n}}\right)^2= \frac{1}{n}\left(1+\sqrt{n}\right)^2.\]

Substituting these back in we obtain

\[\left[ 1+\frac{1}{\sqrt{n}} \right]^{n-2} \geq \frac{\frac{1}{2}(1+\sqrt{n})^2}{\frac{1}{n}\left(1+\sqrt{n}\right)^2}=\tfrac{1}{2}n.\]

#### Approach 2: Manipulate the right hand side to find hidden factors of \((1+1/\sqrt{n})\).

In this less direct (but essentially equivalent) approach we manipulate the right hand side by looking for factors of \((1+1/\sqrt{n})\) that we can cancel from both sides.

For this it’s handy to note that we can write \(n-1=(\sqrt{n}+1)(\sqrt{n}-1)\). Thus \[\begin{align*} \left[ 1+\frac{1}{\sqrt{n}} \right]^n &\geq 1+\sqrt{n}+\frac{n-1}{2}. \\ &= 1+\sqrt{n}+\frac{(\sqrt{n}+1)(\sqrt{n}-1)}{2}. \\ &= (1+\sqrt{n})\left(1+\frac{\sqrt{n}-1}{2}\right). \\ &= (1+\sqrt{n})\left(\frac{\sqrt{n}+1}{2}\right). \\ &= \frac{1}{2}(1+\sqrt{n})^2 \\ \end{align*}\] This still isn’t quite what we want — on the right hand side we have \((1+\sqrt{n})\), when we want \((1+1/\sqrt{n})\) — but we can fix this by noting that \((1+\sqrt{n})=\sqrt{n}(1+1/\sqrt{n})\). Thus \[\begin{align*} \left[ 1+\frac{1}{\sqrt{n}} \right]^n &\geq \frac{1}{2}(1+\sqrt{n})^2 \\ &= \frac{1}{2}n\left[1+\frac{1}{\sqrt{n}}\right]^2, \\ \end{align*}\]which, after dividing through by \((1+1/\sqrt{n})^2\), implies that \[\left[1+\frac{1}{\sqrt{n}}\right]^{n-2} \geq \frac{1}{2}n,\] as desired.

Is it important that \(n>1\)?

You might like to check if the inequality is true at \(n=1\), or \(n=0\). You might then consider what would happen if \(n\) was allowed to take non-integer real values.

Where do issues arise?