Tag: Inequalities

Calculate approximations to log 2

by RoRi

November 7, 2015

Theorem 6.5 (page 240 of Apostol) states that if $0 < x < 1$ and if $m \geq 1$ then

$\log \frac{1+x}{1-x} = 2 \left( x + \frac{x^3}{3} + \cdots + \frac{x^{2m-1}}{2m-1} \right) + R_m (x),$

with

$\frac{x^{2m+1}}{2m+1} < R_m (x) \leq \frac{2-x}{1-x} \frac{x^{2m+1}}{2m+1}.$

Let $m = 5$ and $x = \frac{1}{3}$ and apply this theorem to approximate $\log 2$ with sufficient precision to obtain

$0.6931460 < \log 2 < 0.6931476.$

By the theorem we have

$\log \frac{1+x}{1-x} = 2 \left( x + \frac{x^3}{3} + \cdots + \frac{x^{2m-1}}{2m-1} \right) + R_m (x),$

where

$\frac{x^{2m-1}}{2m-1} < R_m (x) \leq \left( \frac{2-x}{1-x} \right) \left( \frac{x^{2m+1}}{2m+1} \right).$

Therefore, letting $x = \frac{1}{3}$ and $m = 5$ we have

$\log \frac{1+\frac{1}{3}}{1-\frac{1}{3}} = 2 \left( \frac{1}{3} + \frac{(1/3)^3}{3} + \frac{(1/3)^5}{5} + \frac{(1/3)^7}{7} + \frac{(1/3)^9}{9} \right) + R_5 \left( \frac{1}{3} \right).$

Therefore,

$0.693146 < \log 2 < 0.6931476.$

Prove inequalities of the logarithm with respect to some series

by RoRi

November 6, 2015

Consider a partition $P = \{ a_0, a_1, a_2, \ldots, a_n \}$ of the interval $[1,x]$ for some $x > 1$ .

Find step functions that are constant on the open subintervals of $P$ and integrate to derive the inequalities:
$\sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_k} \right) < \log x < \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_{k-1}} \right).$
Give a geometric interpretation of the inequalities in part (a).
Find a particular partition $P$ (i.e., choose particular values for $a_0, a_1, \ldots, a_n$ ) to establish the following inequalities for $n > 1$ ,
$\sum_{k=2}^n \frac{1}{k} < \log n < \sum_{k=1}^{n-1} \frac{1}{k}.$

Proof. We define step function $s$ and $t$ by
$\begin{align*} s(x) &= \frac{1}{a_k} & \text{for } x &\in [a_{k-1}, a_k ) \\ t(x) &= \frac{1}{a_{k-1}} & \text{for } x &\in [a_{k-1}, a_k ). \end{align*}$

Since $\frac{1}{x}$ is strictly decreasing on $\mathbb{R}_{>0}$ , we have

$s(x) < \frac{1}{x} < t(x) \qquad \text{for all } x \in \mathbb{R}_{>0}.$

Therefore, using the definition of the integral of a step function as a sum,

$\begin{align*} && \int_1^x s(u) \, du &< \int_1^x \frac{1}{u} \, du < \int_1^x t(u) \, du \\ \implies && \sum_{k=1} s_k (a_k - a_{k-1} ) &< \log x < \sum_{k=1}^n t_k (a_k - a_{k-1}) \\ \implies && \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_k} \right) &< \log x < \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_{k-1}} \right). \qquad \blacksquare \end{align*}$
Geometrically, these inequalities say that the area under the curve $\frac{1}{x}$ lies between the step functions that take on the values $\frac{1}{a_k}$ and $\frac{1}{a_{k-1}}$ for each $x \in [ a_{k-1}, a_k )$ .
Proof. To establish these inequalities we pick the partition,
$P = \{ 1, 2, 3, \ldots, n \} \qquad \text{for } n > 1.$

Then, applying part (a) we have

$\begin{align*} && \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_k} \right) &< \log x < \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_{k-1}} \right) \\ \implies && \sum_{k=1}^n \frac{1}{a_k} &< \log n < \sum_{k=1}^n \frac{1}{a_{k-1}} &(\text{since } a_k - a_{k-1} = 1) \\ \implies && \sum_{k=2}^n \frac{1}{k} &< \log n < \sum_{k=1}^{n-1} \frac{1}{k}. \end{align*}$

The final line follows since $a_0 = 1, a_1 = 2, \ldots$ so the sum on the left starts with $\frac{1}{2}$ and the sum on the right only runs to $\frac{1}{n-1}$ . These were the inequalities requested $. \qquad \blacksquare$

Prove the inequalities 1 – 1/x < log x < x - 1

by RoRi

November 6, 2015

Define the following functions:
$f(x) = x - 1 - \log x, \qquad g(x) = \log x - 1 + \frac{1}{x},$

for $x > 0$ . Prove that for $x > 0$ and $x \neq 1$ the inequalities

$1 - \frac{1}{x} < \log x < x - 1$

hold. Proceed by examining the signs of the derivatives $f'$ and $g'$ . When $x = 1$ these are equalities.
Draw the graphs of the functions
$A(x) = x - 1, \qquad B(x) = 1 - \frac{1}{x}, \qquad x > 0.$

Interpret the inequalities in part (a) geometrically.

First, we compute the derivatives of the functions,
$\begin{align*} f(x) &= x - 1 - \log x & \implies && f'(x) &= 1 - \frac{1}{x} \\ g(x) &= \log x - 1 + \frac{1}{x} & \implies && g'(x) &= \frac{1}{x} - \frac{1}{x^2} = \frac{1}{x} \left( 1 - \frac{1}{x} \right). \end{align*}$

Then, considering the derivative of $f$ ,

$\begin{align*} f'(x) = 1 - \frac{1}{x} && \implies && f'(x) > 0 \quad &\text{when } x > 1 \\ \text{and} && \implies && f'(x) < 0 \quad &\text{when } 0 < x < 1. \end{align*}$

Therefore, the function $f$ has a minimum at $x = 1$ . Since we can directly evaluate $f(1) = 0$ , this means $f(x) > 0$ for $x > 0$ and $x \neq 1$ . Therefore, $x - 1 > \log x$ .

Next, looking at the derivative of $g$ ,

$\begin{align*} g'(x) = \frac{1}{x} \left( 1 - \frac{1}{x} \right) && \implies && g'(x) > 0 \quad & \text{when } x > 1 \\ \text{and} && \implies && g'(x) < 0 \quad &\text{when } 0 < x < 1. \end{align*}$

Therefore, the function $g$ has a minimum at $x = 1$ . Since $g(1) = 0$ this implies $g(x) > 0$ for $x > 0$ and $x \neq 1$ . Thus,

$1 - \frac{1}{x} < \log x.$

Putting these two pieces together we have established the requested inequalities:

$1 - \frac{1}{x} < \log x < x - 1 \qquad \text{for } x > 0 \text{ and } x \neq 1. \qquad \blacksquare$
The graph of the two functions is:

The inequalities in part (a) imply that the graph of $\log x$ must lie strictly between the graphs of $A(x)$ and $B(x)$ shown above (and so $\log 1 = 0$ ).

Application of the mean-value theorem for integrals

by RoRi

October 23, 2015

Let $\varphi$ be a function with second derivative $\varphi''$ continuous and nonzero on an interval $[a,b]$ . Furthermore, let $m > 0$ be a constant such that
$\varphi'(t) \geq m \qquad \text{for all } t \in [a,b].$

Use the second mean-value theorem for integrals (Theorem 5.5 in Apostol) to prove the inequality

$\left| \int_a^b \sin \varphi(t) \, dt \right| \leq \frac{4}{m}.$
If $a > 0$ prove that
$\left| \int_a^x \sin (t^2) \, dt \right| \leq \frac{2}{a} \qquad \text{for } x > a.$

Proof. Since we have the assumption that $\varphi'(t) \geq m > 0$ for all $t \in [a,b]$ we may divide by $\varphi'(t)$ , to obtain
$\left| \int_a^b \sin \varphi(t) \, dt \right| = \left| \int_a^b \frac{\sin \varphi(t)}{\varphi'(t)} \cdot \varphi'(t) \, dt \right|.$

Then, to apply the second mean-value theorem for integrals (Theorem 5.5 of Apostol) we define functions

$f(t) = \frac{1}{\varphi'(t)} \qquad \text{and} \qquad g(t) = \varphi'(t) \sin \varphi(t).$

The function $g$ is continuous since $\sin \varphi(t)$ is a composition of continuous functions (we know $\varphi(t)$ is continuous since it is differentiable) and $\varphi'(t)$ is continuous (again, it is differentiable since $\varphi''(t)$ exists and is continuous by assumption). Then the product of continuous function is also continuous, which establishes that $g$ is continuous. We also know that $f$ meets the conditions of the theorem since it has derivative given by

$f'(t) = - \frac{\varphi''(t)}{(\varphi'(t))^2}.$

This derivative is continuous since $\varphi''(t)$ and $\varphi'(t)$ are continuous and $\varphi'(t)$ is nonzero. Furthermore, this derivative does not change since on $[a,b]$ since $\varphi''(t)$ is nonzero on $[a,b]$ (and by Bolzano’s theorem we know that a continuous function that changes sign must have a zero). Therefore, we can apply the second mean-value theorem:

$\begin{align*} \left| \int_a^b \frac{\varphi'(t) \sin \varphi(t)}{\varphi'(t)} \, dt \right| &= \left| \frac{1}{\varphi'(a)} \int_a^c \varphi'(t) \sin \varphi(t) \, dt + \frac{1}{\varphi'(b)} \int_c^b \varphi'(t) \sin \varphi(t) \, dt \right| \\ &\leq \left| \frac{1}{m} \int_a^c \varphi'(t) \sin \varphi(t) \, dt + \frac{1}{m} \int_c^b \varphi'(t) \sin \varphi(t) \, dt \right| \\ &\leq \left| -\frac{1}{m} \cos \varphi(t) \Big \rvert_a^c + \left(-\frac{1}{m} \right) \cos \varphi(t) \Big \rvert_c^b \right| \\ & \leq \left| -\frac{1}{m} \cos \varphi(t) \Big \rvert_a^c \right| + \left| -\frac{1}{m} \cos \varphi(t) \Big \rvert_c^b \right| \\ \intertext{(by the Triangle Inequality)} &\leq \left| -\frac{2}{m} \right| + \left| -\frac{2}{m} \right| \\ & = \frac{2}{m} + \frac{2}{m} \\ &= \frac{4}{m}. \qquad \blacksquare \end{align*}$
Proof. Using part (a), we take $\varphi(t) = t^2$ , giving us
$\varphi(t) = t^2 \quad \implies \quad \varphi'(t) = 2t \quad \implies \quad \varphi''(t) = 2.$

Thus, $\varphi''(t)$ is continuous and never changes sign. Furthermore, $\varphi' \geq m = 2a$ (where $a$ is a given constant) and $2a > 0$ since $a> 0$ . Thus,

$\left| \int_a^x \sin (t^2) \, dt \right| \leq \frac{4}{2a} = \frac{2}{a} \qquad \text{for all } x > a. \qquad \blacksquare$

Prove some inequalities using the mean value theorem

by RoRi

September 28, 2015

Prove the following inequalities using the mean value theorem:

$| \sin x - \sin y| \leq |x - y|$ .
$ny^{n-1} (x-y) \leq x^n - y^n \leq nx^{n-1} (x-y) \qquad \text{if } 0 < y \leq x$ , for $n = 1, 2, 3, \ldots$ .

Proof. Define $f(t) = \sin t$ and $g(t) = t$ . Then $f$ and $g$ are continuous and differentiable everywhere so we may apply the mean value theorem. We obtain
$\begin{align*} &&f'(c) (g(x) - g(y)) &= g'(c) (f(x) - f(y)) & (\text{for some } c \in (x,y))\\ \implies && (\cos c) (x-y) &= \sin x - \sin y \\ \implies && | \cos c | | x-y| &= | \sin x - \sin y| \\ \implies && |x-y| &\geq | \sin x - \sin y|. \end{align*}$

The final step follows since $| \cos c | \leq 1$ for all $c. \qquad \blacksquare$
Proof. Let $f(t) = t^n$ , $g(t) = t$ , then $f'(t) = nt^{n-1}$ and $g'(t) = 1$ . So, by the mean-value theorem we have there exists a $c \in (x,y)$ such that,
$\begin{align*} && f'(c) (g(x) - g(y)) &= g'(c)(f(x) - f(y)) \\ \implies && nc^{n-1}(x-y) &= x^n - y^n. \end{align*}$

But, since $x^{n-1}$ is an increasing function on the positive real axis, and we have $0 < y \leq c \leq x$ we know

$y^{n-1} \leq c^{n-1} \leq x^{n-1}.$

Further, since $(x-y) \geq 0$ and $n$ is positive we can multiply all of the terms in the equality by $n (x-y)$ without reversing inequalities to obtain,

$ny^{n-1}(x-y) \leq nc^{n-1} (x-y) \leq nx^{n-1} (x-y).$

Therefore, substituting $nc^{n-1} (x-y) = x^n - y^n$ from above we obtain the requested inequality:

$ny^{n-1} (x-y) \leq x^n - y^n \leq nx^{n-1} (x-y). \qquad \blacksquare$

Give a geometric proof that sin x < x for 0 < x < π/2

by RoRi

August 6, 2015

Consider the following figure:

Rendered by QuickLaTeX.com

Compare the area of the triangle $OAP$ with the area of the sector $OAP$ to prove

$\sin x < x \qquad \text{for} \qquad 0 < x < \frac{\pi}{2}.$

Further, prove,

$| \sin x | < |x| \qquad \text{if} \qquad 0 < |x| < \frac{\pi}{2}.$

Proof. Assume the radius is $r = 1$ . For $0 < x < \frac{\pi}{2}$ we have the triangle $OAP_t$ has base length $a = 1$ and height $b = \sin x$ the area is

$\text{Area}(OAP_t) = \frac{1}{2} ab = \frac{\sin x}{2}.$

The area of the circular sector $OAP_s$ is

$\text{Area}(OAP_s) = \frac{x}{2 \pi} \pi r^2 = \frac{x}{2}.$

Since the area of the triangle $OAP_t$ is less than the area of the sector $OAP_s$ we have,

$\begin{align*} \frac{\sin x }{2} < \frac{x}{2} && \implies && \sin x &< x. \end{align*}$

Then, since $\sin (-x) = - \sin x$ we have for $0 < x < \frac{\pi}{2}$

$\sin x < x \quad \implies \quad -\sin x > -x \quad \implies \quad | \sin x | < |x|$

for $0 < |x| < \frac{\pi}{2}. \qquad \blacksquare$

Use dodecagons to deduce an inequality about π

by RoRi

July 30, 2015

By considering dodecagons inscribed and circumscribed about a unit disk, establish the inequalities

$3 < \pi < 12(2-\sqrt{3}).$

First, we draw some pictures of the situation for reference.

Rendered by QuickLaTeX.com

( Note: I don’t know a way to do this without using trig functions, which haven’t been introduced in the text yet. If you have an alternative approach without them, please leave a comment and let us know about it. )

Since these are dodecagons, the angle at the origin of the circle of each triangular sector is $2 \pi / 12 = \pi/6$ , and the angle of the right triangles formed by splitting each of these sectors in half (shown in the diagrams) is then $\pi/12$ . Then we use the fact that

$\tan \left( \frac{\pi}{12} \right) = 2 - \sqrt{3},$

$\sin \left( \frac{\pi}{12} \right) = \frac{\sqrt{3} - 1}{2 \sqrt{2}},$

$\cos \left( \frac{\pi}{12} \right) = \frac{\sqrt{3} + 1}{2 \sqrt{2}}.$

Now, for the circumscribed dodecagon we have the area of the right triangle $T$ with base 1 in the diagram on the left given by

$a(T) = \frac{1}{2} bh = \frac{1}{2} \cdot 1 \cdot (2 - \sqrt{3}) = 1 - \frac{\sqrt{3}}{2}.$

Since there are 24 such triangles in the dodecahedron, we then have the area of the circumscribed dodecahedron $D_c$ given by

$a(D_c) = 24 \left(1 - \frac{\sqrt{3}}{2} \right) = 12 (2 - \sqrt{3}).$

For the inscribed dodecagon we consider the right triangle $T$ with hypotenuse 1 in the diagram. The length of one of the legs is then given by $\sin \left( \frac{\pi}{12} \right) = \frac{\sqrt{3} - 1}{2}$ and the other is given by $\cos \left( \frac{\pi}{12} \right)$ . So the area of the triangle is

$a(T) = \frac{1}{2} bh = \frac{1}{2} \cdot \frac{\sqrt{3}-1}{2 \sqrt{2}} \cdot \frac{\sqrt{3}+1}{2 \sqrt{2}} = \frac{2}{16} = \frac{1}{8}.$

Since there are 24 such triangles in the inscribed dodecahedron, $D_{i}$ we then have,

$a(D_i) = 24 \cdot \frac{1}{8} = 3.$

Since the area of the unit circle is, by definition, $\pi$ , and it lies in between these two dodecahedrons, we have,

$3 < \pi < 12(2 - \sqrt{3}). \qquad \blacksquare$

Prove that (1-a)(1-b)(1-c) >= 8abc if a+b+c=1

by RoRi

July 11, 2015

If $a,b,c \in \mathbb{R}_{>0}$ and $a+b+c = 1$ , prove that

$(1-a)(1-b)(1-c) \geq 8abc.$

Proof. Recall the definition of the $p$ th power mean, $M_p$ ,

$M_p = \left( \frac{x_1^p + \cdots + x_n^p}{n} \right)^{1/p}$

for $x_1, \ldots, x_n$ are $n$ positive real numbers and $p$ is any integer.
By the generalization of the Arithmetic Mean-Geometric Mean inequality (proved in part b of the linked exercise) we know that $M_{-1} \leq M_1$ .

So, for the problem at hand, we consider $M_{-1} (a,b,c) \leq M_1 (a,b,c)$ gives us,

$\left( \frac{a^{-1} + b^{-1} + c^{-1}}{3} \right)^{-1} &\leq \left( \frac{a+b+c}{3} \right).$

Simplifying this expression, we have

$\begin{align*} && \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} &\leq \frac{1}{3} & (a+b+c = 1)\\ \implies && 9 &\leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \\ \implies && 9abc &\leq bc + ac + ab & (\text{multiplying by } abc)\\ \implies && 8abc &\leq bc + ac + ab - abc \\ \implies && 8abc & \leq 1 - (a+b+c) + ab + ac + bc - abc \\ \intertext{where we are just adding 0 to the right hand side since $1 = a+b+c$ implies $1-(a+b+c) = 0$. Then since $1 - (a + b + c) + ab + ac + bc - abc = (1-a)(1-b)(1-c)$ (which we can check by multiplying out the right hand side), we have} \implies && 8abc &\leq (1-a)(1-b)(1-c). \qquad \blacksquare \end{align*}$

Prove the product of sums of n numbers and their reciprocals is greater than n^2

by RoRi

July 11, 2015

Let $x_1, \ldots, x_n \in \mathbb{R}_{>0}$ and let $y_k = \frac{1}{x_k}$ for $k = 1, \ldots, n$ . Prove

$\left( \sum_{k=1}^n x_k \right) \left( \sum_{k=1}^n y_k \right) \geq n^2.$

Proof. First,

$y_k = \frac{1}{x_k} \quad \implies \quad \sqrt{y_k} = \frac{1}{\sqrt{x_k}} \quad \implies \quad \sqrt{x_k} \sqrt{y_k} = 1$

for all $k = 1, \ldots, n$ . We then apply the Cauchy-Schwarz inequality to the numbers $\sqrt{x_1}, \ldots, \sqrt{x_n}$ and $\sqrt{y_1}, \ldots, \sqrt{y_n}$ . So, Cauchy-Schwarz gives us,

$\begin{align*} &&\left( \sum_{k=1}^n a_k b_k \right)^2 &\leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)\\ \implies && \left( \sum_{k=1}^n \sqrt{x_k} \sqrt{y_k} \right)^2 &\leq \left( \sum_{k=1}^n (\sqrt{x_k})^2 \right) \left( \sum_{k=1}^n (\sqrt{y_k})^2 \right) \\ \implies && \left( \sum_{k=1}^n 1 \right)^2 &\leq \left( \sum_{k=1}^n x_k \right) \left( \sum_{k=1}^n y_k \right)\\ \implies && n^2 &\leq \left(\sum_{k=1}^n x_k \right) \left( \sum_{k=1}^n y_k \right). \qquad \blacksquare \end{align*}$

Prove the arithmetic mean – geometric mean inequality

by RoRi

July 10, 2015

We define the geometric mean $G$ of real numbers $x_1, \ldots, x_n$ by

$G := (x_1 \cdot \cdots \cdot x_n)^{1/n}.$

We also recall the definition of the $p$ th power mean, $M_p$ :

$M_p = \left( \frac{x_1^p + \cdots + x_n^p}{n} \right)^{1/p}.$

Prove the arithmetic mean – geometric mean inequality, i.e., prove $G \leq M_1$ where $M_1$ is the $p$ th power mean with $p=1$ (also known as the arithmetic mean).
For integers $p,q$ with $q < 0 < p$ , prove that $M_q < G < M_p$ if $x_1, \ldots, x_n$ are not all equal.

Proof. First, if $x_1 = \cdots = x_n$ , then

$G = (x_1 \cdot \cdots \cdot x_n)^{1/n} = (x_1^n)^{1/n} = x_1 \qquad \text{and} \qquad M_1 = \left( \frac{x_1 + \cdots + x_n}{n} \right) = \left( \frac{nx_1}{n} \right) = x_1.$

Hence, $G = M_1$ for the case $x_1 = \cdots = x_n$ .
Now, if $x_1, \ldots, x_n$ are not all equal then first we write,

$G^n = \left( (x_1 \cdot \cdots \cdot x_n)^{1/n}\right)^n = x_1 \cdot \cdots \cdot x_n.$

Then using the previous exercise, we proceed,

$\begin{align*} &&\left( \frac{1}{G^n} \right) (x_1 \cdot \cdots \cdot x_n) &= 1 \\ \implies && \left( \frac{x_1}{G} \right) \left( \frac{x_2}{G} \right) \cdot \cdots \cdot \left( \frac{x_n}{G} \right) &= 1 \\ \implies && \frac{x_1}{G} + \frac{x_2}{G} + \cdots + \frac{x_n}{G} &> n & (\text{previous exercise})\\ \implies && x_1 + x_2 + \cdots + x_n &> n \cdot G \\ \implies && M_1 &> G. \end{align*}$

Therefore, if $x_1, \ldots, x_n$ are not all equal then we have the strict inequality $G < M_1$ , as requested $. \qquad \blacksquare$
Proof. We’ll start with the inequality on the right first. So, we want to show $G < M_p$ for any $n$ positive real numbers $x_1, \ldots, x_n$ not all equal, where $p$ is a positive integer. First, we’ll want to observe that if $x_1, \ldots, x_n$ are positive real numbers, not all equal, then $x_1^p, \ldots, x_n^p$ are also positive real numbers, not all equal. So from the definition of $M_p$ and letting $M_p (x_1^p, \ldots, x_n^p)$ denote the $p$ th power mean of the numbers $x_1^p, \ldots, x_n^p$ , we have

$\begin{align*} M_1 (x_1^p, \ldots, x_n^p) &= \left( \frac{x_1^p + \cdots + x_n^p}{n} \right) \\ &= \left( \left( \frac{x_1^p + \cdots + x_n^p}{n} \right)^{1/p} \right)^p \\ &= (M_p(x_1, \ldots, x_n))^p. \end{align*}$

So, the observation is that $(M_p (x_1, \ldots, x_n))^p = M_1(x_1^p, \ldots, x_n^p)$ . Now, using part (a), we have,

$\begin{align*} G(x_1, \ldots, x_n)^p &= (x_1 \cdots x_n)^{p/n}\\ &= (x_1^p \cdots x_n^p)^{1/n} \\ &= G(x_1^p, \ldots, x_n^p) \\ &< M_1 (x_1^p, \ldots, x_n^p) & (\text{part (a)})\\ &= (M_p (x_1, \ldots, x_n))^p. \end{align*}$

Hence, for any positive real numbers $x_1, \ldots, x_n$ , not all equal we have $(G(x_1, \ldots, x_n))^p < (M_p (x_1, \ldots, x_n))^p$ which implies (see this exercise) $G < M_p$ . This gives us the inequality on the right.
Now, for $q$ a negative integer, we must show $M_q < G$ . Since $q < 0$ we know that $-q>0$ and so $G < M_{-q}$ from the inequality we just proved. So,

$G^{-q} < (M_{-q})^{-q} \quad \implies \quad G^q > M_q^q \quad \implies \quad G > M_q.$

Where we have used the same exercise again, and the fact that $G^{-q} = \frac{1}{G^q}. \qquad \blacksquare$