## AM-GM Inequality (Part 2)

In Part 1 of this post, I introduced the AM-GM Inequality and showed a few examples of how to apply it to prove certain inequalities and find the minimum value of certain functions. In Part 2, I will start off with the proof (which I skipped in Part 1) and then proceed to a few more miscellaneous examples of clever applications. If you're not interested in the proof, then you can skip the next section and go straight to the examples below.

\dfrac{x_1 + x_2 + \dotsb + x_n}{n} \geq \sqrt[n]{x_1 x_2 \dotsm x_n}

$$ with equality if and only if $x_1 = x_2 = \dotsb = x_n$.

Now suppose we have $n+1$ non-negative real numbers $x_1, x_2, \dotsc , x_n, x_{n+1}$, and assume that for any $n$ non-negative real numbers, the AM-GM inequality holds. For notational simplicity, define $\alpha$ to be the arithmetic mean of the $n+1$ numbers, $$

\alpha = \dfrac{x_1 + x_2 + \dotsb + x_n + x_{n+1}}{n+1}

$$ which can be re-arranged as $$

(n+1) \alpha = x_1 + x_2 + \dotsb + x_n + x_{n+1} \tag{$\star$}

$$ If each of the $n+1$ numbers is equal to $\alpha$, then the AM and GM are both $\alpha$, so there is nothing to prove. Otherwise, there must be at least one of the $x_i$'s that is greater than $\alpha$ and at least one that is less than $\alpha$. Let's say $x_n > \alpha$ and $x_{n+1} < \alpha$ (if this isn't the case, rearrange the labels so that it is). So we know that $$

(x_n - \alpha)(\alpha - x_{n+1}) > 0 \tag{$\star \star$}

$$ a fact which we will use later.

Now define $y = x_n + x_{n+1} - \alpha$. Note that $y$ is positive since it is greater than $x_n - \alpha$. Furthermore, the AM of the $n$ numbers $x_1, x_2, \dotsc , x_{n-1}, y$ is $\alpha$: $$

\begin{align}

\dfrac{x_1 + x_2 + \dotsb + x_{n-1} + y}{n} &= \dfrac{x_1 + x_2 + \dotsb + x_{n-1} + (x_n + x_{n+1} - \alpha)}{n} \\[3mm]

&= \dfrac{(x_1 + x_2 + \dotsb + x_{n-1} + x_n + x_{n+1}) - \alpha}{n} \\[3mm]

&= \dfrac{(n+1)\alpha - \alpha}{n} \ \ \ \text{[by $\star$]} \\[3mm]

&= \dfrac{n \alpha}{n} \\[2mm]

&= \alpha

\end{align}

$$ By the induction hypothesis, the AM-GM inequality holds for the $n$ numbers $x_1, x_2, \dotsc , x_{n-1}, y$, i.e. $$

\begin{align}

&&\sqrt[n]{x_1 x_2 \dotsm x_{n-1} y} \ &\leq \ \alpha& \\

\implies &&x_1 x_2 \dotsm x_{n-1} y \ &\leq \ \alpha^n& \\

\implies &&x_1 x_2 \dotsm x_{n-1} y \cdot \alpha \ &\leq \ \alpha^{n+1}& \tag{$\spadesuit$}

\end{align}

$$ The last ingredient is the following: $$

\begin{align}

(x_n - \alpha)(\alpha - x_{n+1})

= \ &\underbrace{(x_n + x_{n+1} - \alpha)}_{y} \alpha - x_n x_{n+1} \\[2mm]

> \ &0 \ \ \ \text{[by $\star \star$]} \\[2mm]

\implies y \alpha > \ &x_n x_{n+1}

\end{align}

$$ Plugging this into $( \spadesuit )$ yields $$

\begin{align}

\alpha^{n+1} &\geq x_1 x_2 \dotsm x_{n-1} (y \alpha) \\[2mm]

&> x_1 x_2 \dotsm x_{n-1} (x_n x_{n+1}) \\[2mm]

\implies \alpha &> \sqrt[n+1]{x_1 x_2 \dotsm x_{n-1} x_n x_{n+1}}

\end{align}

$$ i.e. the arithmetic mean is greater than the geometric mean. This completes the proof.

$\square$

In Part 1, I showed one easy example and two examples where we could apply the AM-GM inequality to find the minimum value of a function subject to a volume-like constraint. In this part, I'll show 3 more examples of clever applications, in order of increasing difficulty.

\dfrac{a_1}{b_1} + \dfrac{a_2}{b_2} + \dotsb + \dfrac{a_n}{b_n} \geq n

$$

\dfrac{1}{n} \left[ \dfrac{a_1}{b_1} + \dfrac{a_2}{b_2} + \dotsb + \dfrac{a_n}{b_n} \right]

\geq

\sqrt[n \uproot4]{\dfrac{a_1}{b_1} \dfrac{a_2}{b_2} \dotsm \dfrac{a_n}{b_n}}

= 1

$$ where the last equality is true since the $a_i$'s and $b_i$'s are the same list of numbers, just rearranged. Multiplying both sides by $n$ yields the desired inequality.

$\square$

In Example 4, there was no volume-like constraint on the values of the $a_i$'s, but we were able to produce one on the fractions $\frac{a_i}{b_i}$ by multiplying them all together. In Example 5, we can do something similar- this one is admittedly

\begin{gather}

x_1 + \dfrac{1}{x_2} = 4 \\[3mm]

x_2 + \dfrac{1}{x_3} = 1 \\[3mm]

\vdots \tag{$\diamondsuit$}\\[3mm]

x_{99} + \dfrac{1}{x_{100}} = 4 \\[3mm]

x_{100} + \dfrac{1}{x_1} = 1

\end{gather}

$$

\begin{gather}

x_1 + \dfrac{1}{x_2} \geq 2 \sqrt{\dfrac{x_1}{x_2}} \\[3mm]

x_2 + \dfrac{1}{x_3} \geq 2 \sqrt{\dfrac{x_2}{x_3}} \\[3mm]

\vdots \tag{$\clubsuit$} \\[3mm]

x_{100} + \dfrac{1}{x_1} \geq 2 \sqrt{\dfrac{x_{100}}{x_1}}

\end{gather}

$$ Note that putting the $x$'s under square roots is valid since we are only considering positive solutions. Now, both sides of $( \clubsuit )$ are positive in each of these inequalities, so we can multiply them together without reversing the direction of the inequality: $$

\begin{align}

\left( x_1 + \dfrac{1}{x_2} \right)

\left( x_2 + \dfrac{1}{x_3} \right)

\dotsm

\left( x_{100} + \dfrac{1}{x_1} \right)

&\geq 2^{100} \sqrt{\dfrac{x_1}{x_2} \dfrac{x_2}{x_3} \dotsm \dfrac{x_{100}}{x_1}} \\[2mm]

&= 2^{100} \cdot 1 \\[2mm]

&= 2^{100}

\end{align}

$$ thus $$

\left( x_1 + \dfrac{1}{x_2} \right)

\left( x_2 + \dfrac{1}{x_3} \right)

\dotsm

\left( x_{100} + \dfrac{1}{x_1} \right)

\geq 2^{100} \tag{$\heartsuit$}

$$ On the other hand, multiplying together the original system of equations $( \diamondsuit )$ shows that the product on the left side of $( \heartsuit )$ is equal to $4^{50} \cdot 1^{50} = 2^{100}$. In other words, the inequality $( \heartsuit )$ is an exact equality. This implies, in turn, that each of the inequalities in $( \clubsuit )$ are exact equalities too, since otherwise $( \heartsuit )$ would not be an exact equality.

Finally, we can solve for the variables: $$

\begin{align}

&& x_1 + \dfrac{1}{x_2} &= 2 \sqrt{\dfrac{x_1}{x_2}} \\[3mm]

&\implies& x_1 - 2 \sqrt{\dfrac{x_1}{x_2}} + \dfrac{1}{x_2} &= 0 \\[3mm]

&\implies& \left( \sqrt{x_1} - \sqrt{\dfrac{1}{x_2}} \right)^2 &= 0 \\[3mm]

&\implies& \sqrt{x_1} &= \sqrt{\dfrac{1}{x_2}} \\[3mm]

&\implies& x_1 &= \dfrac{1}{x_2}

\end{align}

$$ Similarly, $x_2 = \frac{1}{x_3}, \dotsc, x_{100} = \frac{1}{x_1}$. These, combined with the original system of equations $( \diamondsuit )$, show that the solution is $x_1 = 2, x_2 = \frac{1}{2}, x_3 = 2, x_4 = \frac{1}{2}, \dotsc, x_{99} = 2, x_{100} = \frac{1}{2}$.

$\square$

The sixth and final example will require proving a lemma, but I thought this example (and the lemma itself) was so neat that it would be worth the effort.

r_1^2 + r_2^2 + \dotsb + r_n^2 = \left( \dfrac{a_{n-1}}{a_n} \right)^2 - 2 \dfrac{a_{n-2}}{a_n}

$$

p(x) = a_n (x-r_1)(x-r_2) \dotsm (x-r_n) \tag{$\dagger$}

$$ When expanding this product, we add terms consisting of $\pm a_n$ times $n$ items, each either an $x$ or one of the $r_i$'s, then combine "like terms," i.e. those with the same power of $x$. For $0 \leq k \leq n$, a term in the expansion contributes to the $x^k$ coefficient when, from the $n$ factors in the product $( \dagger )$, we choose an $x$ from $k$ of them and the $-r_i$ from the other $n-k$. Therefore, the coefficient of $x^k$ in the expansion is $$

a_n (-1)^{n-k} \sum_{1 \leq i_1 < i_2 < \dotsb < i_{n-k} \leq n}{r_{i_1} r_{i_2} \dotsm r_{i_{n-k}}} \tag{$\ddagger$}

$$ Since the product $( \dagger )$ is an equivalent way of writing $p(x)$, the sum $( \ddagger )$ must be equal to the coefficient $a_k$.

On the other hand, the sum of the squares of the roots can be written as $$

\begin{align}

r_1^2 + r_2^2 + \dotsb + r_n^2

&= (r_1 + r_2 + \dotsb + r_n)^2 - 2 \sum_{1 \leq i_1 < i_2 \leq n}{r_{i_1} r_{i_2}} \\[3mm]

&= \left( \dfrac{1}{a_n (-1)^{n-(n-1)}}a_{n-1} \right)^2

-2 \dfrac{1}{a_n (-1)^{n-(n-2)}}a_{n-2} \tag{$\maltese$} \\[3mm]

&= \left( \dfrac{a_{n-1}}{a_n} \right)^2 - 2 \dfrac{a_{n-2}}{a_n}

\end{align}

$$ where the equality $( \maltese )$ was obtained by plugging in $(n-1)$ and $(n-2)$ in for $k$ in $( \ddagger )$. This completes the proof.

$\square$

With the lemma and a slick AM-GM application, the final example will be a piece of cake.

By the lemma (with $a_n = 1$), the sum of the squares of the roots is $a_{n-1}^2 - 2 a_{n-2}$. Also, plugging in $k=n$ in $( \ddagger )$ tells us that the product of the squares of the roots is $a_0^2$.

By the AM-GM inequality, we have $$

\begin{align}

&&\dfrac{r_1^2 + r_2^2 + \dotsb + r_n^2}{n} &\geq \sqrt[n \uproot2]{r_1^2 r_2^2 \dotsm r_n^2} \\[3mm]

&\implies& \dfrac{a_{n-1}^2 - 2 a_{n-2}}{n} &\geq \sqrt[n]{a_0^2} \\[3mm]

&\implies& \dfrac{1 \pm 2}{n} &\geq 1 \ \ \ [\text{since all coefficients are }\pm 1 ] \\[3mm]

&\implies& 3 &\geq n

\end{align}

$$ $\square$

That will do it for this post. Please post any questions in the comments section. Thanks for reading, and stay tuned for part 3...

#### Proof of the AM-GM Inequality

**AM-GM Inequality:**For $x_1, x_2, \dotsc , x_n \geq 0$, $$\dfrac{x_1 + x_2 + \dotsb + x_n}{n} \geq \sqrt[n]{x_1 x_2 \dotsm x_n}

$$ with equality if and only if $x_1 = x_2 = \dotsb = x_n$.

*Proof:*The proof is by induction. The base case for induction is $n=1$, for which it's obvious that the inequality holds (in fact, it's an equality).Now suppose we have $n+1$ non-negative real numbers $x_1, x_2, \dotsc , x_n, x_{n+1}$, and assume that for any $n$ non-negative real numbers, the AM-GM inequality holds. For notational simplicity, define $\alpha$ to be the arithmetic mean of the $n+1$ numbers, $$

\alpha = \dfrac{x_1 + x_2 + \dotsb + x_n + x_{n+1}}{n+1}

$$ which can be re-arranged as $$

(n+1) \alpha = x_1 + x_2 + \dotsb + x_n + x_{n+1} \tag{$\star$}

$$ If each of the $n+1$ numbers is equal to $\alpha$, then the AM and GM are both $\alpha$, so there is nothing to prove. Otherwise, there must be at least one of the $x_i$'s that is greater than $\alpha$ and at least one that is less than $\alpha$. Let's say $x_n > \alpha$ and $x_{n+1} < \alpha$ (if this isn't the case, rearrange the labels so that it is). So we know that $$

(x_n - \alpha)(\alpha - x_{n+1}) > 0 \tag{$\star \star$}

$$ a fact which we will use later.

Now define $y = x_n + x_{n+1} - \alpha$. Note that $y$ is positive since it is greater than $x_n - \alpha$. Furthermore, the AM of the $n$ numbers $x_1, x_2, \dotsc , x_{n-1}, y$ is $\alpha$: $$

\begin{align}

\dfrac{x_1 + x_2 + \dotsb + x_{n-1} + y}{n} &= \dfrac{x_1 + x_2 + \dotsb + x_{n-1} + (x_n + x_{n+1} - \alpha)}{n} \\[3mm]

&= \dfrac{(x_1 + x_2 + \dotsb + x_{n-1} + x_n + x_{n+1}) - \alpha}{n} \\[3mm]

&= \dfrac{(n+1)\alpha - \alpha}{n} \ \ \ \text{[by $\star$]} \\[3mm]

&= \dfrac{n \alpha}{n} \\[2mm]

&= \alpha

\end{align}

$$ By the induction hypothesis, the AM-GM inequality holds for the $n$ numbers $x_1, x_2, \dotsc , x_{n-1}, y$, i.e. $$

\begin{align}

&&\sqrt[n]{x_1 x_2 \dotsm x_{n-1} y} \ &\leq \ \alpha& \\

\implies &&x_1 x_2 \dotsm x_{n-1} y \ &\leq \ \alpha^n& \\

\implies &&x_1 x_2 \dotsm x_{n-1} y \cdot \alpha \ &\leq \ \alpha^{n+1}& \tag{$\spadesuit$}

\end{align}

$$ The last ingredient is the following: $$

\begin{align}

(x_n - \alpha)(\alpha - x_{n+1})

= \ &\underbrace{(x_n + x_{n+1} - \alpha)}_{y} \alpha - x_n x_{n+1} \\[2mm]

> \ &0 \ \ \ \text{[by $\star \star$]} \\[2mm]

\implies y \alpha > \ &x_n x_{n+1}

\end{align}

$$ Plugging this into $( \spadesuit )$ yields $$

\begin{align}

\alpha^{n+1} &\geq x_1 x_2 \dotsm x_{n-1} (y \alpha) \\[2mm]

&> x_1 x_2 \dotsm x_{n-1} (x_n x_{n+1}) \\[2mm]

\implies \alpha &> \sqrt[n+1]{x_1 x_2 \dotsm x_{n-1} x_n x_{n+1}}

\end{align}

$$ i.e. the arithmetic mean is greater than the geometric mean. This completes the proof.

$\square$

#### Some more examples of AM-GM applications

In Part 1, I showed one easy example and two examples where we could apply the AM-GM inequality to find the minimum value of a function subject to a volume-like constraint. In this part, I'll show 3 more examples of clever applications, in order of increasing difficulty.

**Example 4:**Let $a_1, a_2, \dotsc , a_n$ be a sequence of positive numbers and $b_1, b_2, \dotsc , b_n$ be a permutation of the $a_i$'s. Show that $$\dfrac{a_1}{b_1} + \dfrac{a_2}{b_2} + \dotsb + \dfrac{a_n}{b_n} \geq n

$$

*Proof*: The AM-GM inequality implies $$\dfrac{1}{n} \left[ \dfrac{a_1}{b_1} + \dfrac{a_2}{b_2} + \dotsb + \dfrac{a_n}{b_n} \right]

\geq

\sqrt[n \uproot4]{\dfrac{a_1}{b_1} \dfrac{a_2}{b_2} \dotsm \dfrac{a_n}{b_n}}

= 1

$$ where the last equality is true since the $a_i$'s and $b_i$'s are the same list of numbers, just rearranged. Multiplying both sides by $n$ yields the desired inequality.

$\square$

In Example 4, there was no volume-like constraint on the values of the $a_i$'s, but we were able to produce one on the fractions $\frac{a_i}{b_i}$ by multiplying them all together. In Example 5, we can do something similar- this one is admittedly

*shamelessly*set up for AM-GM application, but I still thought it was cool enough to show.**Example 5:**Find the positive solutions to the system of equations: $$\begin{gather}

x_1 + \dfrac{1}{x_2} = 4 \\[3mm]

x_2 + \dfrac{1}{x_3} = 1 \\[3mm]

\vdots \tag{$\diamondsuit$}\\[3mm]

x_{99} + \dfrac{1}{x_{100}} = 4 \\[3mm]

x_{100} + \dfrac{1}{x_1} = 1

\end{gather}

$$

*Solution*: The sum $x_1 + \frac{1}{x_2}$ is 2 times the arithmetic mean of the numbers $x_1$ and $\frac{1}{x_2}$, so by the AM-GM inequality, we have (using the same logic on the other 98 variables as well): $$\begin{gather}

x_1 + \dfrac{1}{x_2} \geq 2 \sqrt{\dfrac{x_1}{x_2}} \\[3mm]

x_2 + \dfrac{1}{x_3} \geq 2 \sqrt{\dfrac{x_2}{x_3}} \\[3mm]

\vdots \tag{$\clubsuit$} \\[3mm]

x_{100} + \dfrac{1}{x_1} \geq 2 \sqrt{\dfrac{x_{100}}{x_1}}

\end{gather}

$$ Note that putting the $x$'s under square roots is valid since we are only considering positive solutions. Now, both sides of $( \clubsuit )$ are positive in each of these inequalities, so we can multiply them together without reversing the direction of the inequality: $$

\begin{align}

\left( x_1 + \dfrac{1}{x_2} \right)

\left( x_2 + \dfrac{1}{x_3} \right)

\dotsm

\left( x_{100} + \dfrac{1}{x_1} \right)

&\geq 2^{100} \sqrt{\dfrac{x_1}{x_2} \dfrac{x_2}{x_3} \dotsm \dfrac{x_{100}}{x_1}} \\[2mm]

&= 2^{100} \cdot 1 \\[2mm]

&= 2^{100}

\end{align}

$$ thus $$

\left( x_1 + \dfrac{1}{x_2} \right)

\left( x_2 + \dfrac{1}{x_3} \right)

\dotsm

\left( x_{100} + \dfrac{1}{x_1} \right)

\geq 2^{100} \tag{$\heartsuit$}

$$ On the other hand, multiplying together the original system of equations $( \diamondsuit )$ shows that the product on the left side of $( \heartsuit )$ is equal to $4^{50} \cdot 1^{50} = 2^{100}$. In other words, the inequality $( \heartsuit )$ is an exact equality. This implies, in turn, that each of the inequalities in $( \clubsuit )$ are exact equalities too, since otherwise $( \heartsuit )$ would not be an exact equality.

Finally, we can solve for the variables: $$

\begin{align}

&& x_1 + \dfrac{1}{x_2} &= 2 \sqrt{\dfrac{x_1}{x_2}} \\[3mm]

&\implies& x_1 - 2 \sqrt{\dfrac{x_1}{x_2}} + \dfrac{1}{x_2} &= 0 \\[3mm]

&\implies& \left( \sqrt{x_1} - \sqrt{\dfrac{1}{x_2}} \right)^2 &= 0 \\[3mm]

&\implies& \sqrt{x_1} &= \sqrt{\dfrac{1}{x_2}} \\[3mm]

&\implies& x_1 &= \dfrac{1}{x_2}

\end{align}

$$ Similarly, $x_2 = \frac{1}{x_3}, \dotsc, x_{100} = \frac{1}{x_1}$. These, combined with the original system of equations $( \diamondsuit )$, show that the solution is $x_1 = 2, x_2 = \frac{1}{2}, x_3 = 2, x_4 = \frac{1}{2}, \dotsc, x_{99} = 2, x_{100} = \frac{1}{2}$.

$\square$

The sixth and final example will require proving a lemma, but I thought this example (and the lemma itself) was so neat that it would be worth the effort.

**Lemma:**Let $p(x) = a_n x^n + a_{n-1} x^{n-1} + \dotsb + a_1 x + a_0$ be a polynomial in the variable $x$ (could be a real or complex variable and real or complex coefficients), and let $r_1, r_2, \dotsc, r_n$ be the roots of $p(x)$ (in general, these are complex numbers, and there may be repeats). Then the sum of the squares of the roots is related to the coefficients by the following formula: $$r_1^2 + r_2^2 + \dotsb + r_n^2 = \left( \dfrac{a_{n-1}}{a_n} \right)^2 - 2 \dfrac{a_{n-2}}{a_n}

$$

*Proof*: The polynomial $p(x)$ can alternatively be written as $$p(x) = a_n (x-r_1)(x-r_2) \dotsm (x-r_n) \tag{$\dagger$}

$$ When expanding this product, we add terms consisting of $\pm a_n$ times $n$ items, each either an $x$ or one of the $r_i$'s, then combine "like terms," i.e. those with the same power of $x$. For $0 \leq k \leq n$, a term in the expansion contributes to the $x^k$ coefficient when, from the $n$ factors in the product $( \dagger )$, we choose an $x$ from $k$ of them and the $-r_i$ from the other $n-k$. Therefore, the coefficient of $x^k$ in the expansion is $$

a_n (-1)^{n-k} \sum_{1 \leq i_1 < i_2 < \dotsb < i_{n-k} \leq n}{r_{i_1} r_{i_2} \dotsm r_{i_{n-k}}} \tag{$\ddagger$}

$$ Since the product $( \dagger )$ is an equivalent way of writing $p(x)$, the sum $( \ddagger )$ must be equal to the coefficient $a_k$.

On the other hand, the sum of the squares of the roots can be written as $$

\begin{align}

r_1^2 + r_2^2 + \dotsb + r_n^2

&= (r_1 + r_2 + \dotsb + r_n)^2 - 2 \sum_{1 \leq i_1 < i_2 \leq n}{r_{i_1} r_{i_2}} \\[3mm]

&= \left( \dfrac{1}{a_n (-1)^{n-(n-1)}}a_{n-1} \right)^2

-2 \dfrac{1}{a_n (-1)^{n-(n-2)}}a_{n-2} \tag{$\maltese$} \\[3mm]

&= \left( \dfrac{a_{n-1}}{a_n} \right)^2 - 2 \dfrac{a_{n-2}}{a_n}

\end{align}

$$ where the equality $( \maltese )$ was obtained by plugging in $(n-1)$ and $(n-2)$ in for $k$ in $( \ddagger )$. This completes the proof.

$\square$

**Note:**the result in the lemma is one of**Newton's identities**(or**Newton sums**) for polynomials, and the formulas $( \ddagger )$ (one formula for each value of $k$) are called**Vieta's formulas**. There are similar Newton's identities for the sum of the third, fourth, fifth, etc. powers of the roots. The formulas are recursive, each depending on the lower-degree Newton sums.With the lemma and a slick AM-GM application, the final example will be a piece of cake.

**Example 6:**A polynomial with all coefficients equal to $\pm 1$ and only real roots has degree at most 3.*Proof*: Suppose we have a polynomial $p(x) = a_n x^n + a_{n-1} x^{n-1} + \dotsb + a_1 x + a_0$ where each of the $a_i$'s is either $1$ or $-1$ and with roots $r_1, r_2, \dotsc, r_n$. We can assume that the leading coefficient $a_n$ is $+1$ since the polynomials with $a_n = -1$ are just the negatives of polynomials with $a_n = +1$.By the lemma (with $a_n = 1$), the sum of the squares of the roots is $a_{n-1}^2 - 2 a_{n-2}$. Also, plugging in $k=n$ in $( \ddagger )$ tells us that the product of the squares of the roots is $a_0^2$.

By the AM-GM inequality, we have $$

\begin{align}

&&\dfrac{r_1^2 + r_2^2 + \dotsb + r_n^2}{n} &\geq \sqrt[n \uproot2]{r_1^2 r_2^2 \dotsm r_n^2} \\[3mm]

&\implies& \dfrac{a_{n-1}^2 - 2 a_{n-2}}{n} &\geq \sqrt[n]{a_0^2} \\[3mm]

&\implies& \dfrac{1 \pm 2}{n} &\geq 1 \ \ \ [\text{since all coefficients are }\pm 1 ] \\[3mm]

&\implies& 3 &\geq n

\end{align}

$$ $\square$

That will do it for this post. Please post any questions in the comments section. Thanks for reading, and stay tuned for part 3...