Proofing question (1 Viewer)

AbstractBlade · Mar 28, 2021

Question 25.ii)

CM_Tutor · Mar 28, 2021

The $n=2$ case of the AM-GM-HM inequality states that, so long as $a, b > 0$

$\frac{a+b}{2}\geqslant \sqrt{ab} \geqslant \cfrac{2}{\cfrac{1}{a}+\cfrac{1}{b}}$

It can be proven easily and every MX2 student should be able to do this from memory:

Let $x$ and $y$ be two positive real numbers. Since they must then have positive square roots and all non-zero reals have a positive square, it follows that:

AM-GM:

$\begin{align*} \left(\sqrt{x}-\sqrt{y}\right)^2 &\geqslant 0 \\ x - 2\sqrt{xy} + y &\geqslant 0 \\ \frac{x+ y}{2} &\geqslant \sqrt{xy} \end{align*}$

GM-HM:

$\begin{align*} \left(\sqrt{\frac{1}{x}}-\sqrt{\frac{1}{y}}\right)^2 &\geqslant 0 \\ \frac{1}{x} - 2\sqrt{\frac{1}{x}}\sqrt{\frac{1}{y}} + \frac{1}{y} &\geqslant 0 \\ \frac{1}{x}+ \frac{1}{y} &\geqslant \frac{2}{\sqrt{xy}} \\ \sqrt{xy} &\geqslant \cfrac{2}{\cfrac{1}{x}+ \cfrac{1}{y}} \end{align*}$

Combining these gives the full AM-GM-HM inequality for $n=2$ .

CM_Tutor · Mar 28, 2021

Question 25(i)

If $a+b=6$ , and $a, b > 0$ , show that $\frac{1}{a} + \frac{1}{b} \geqslant \frac{2}{3}$

Solution using AM-HM:

Let $x=a$ and $y=b$ in the statements in the previous post:

$\begin{align*} \frac{x+y}{2} &\geqslant \cfrac{2}{\cfrac{1}{x} + \cfrac{1}{y}} \\ \frac{a+b}{2} &\geqslant \cfrac{2}{\cfrac{1}{a} + \cfrac{1}{b}} \qquad \qquad \text{using $x=a$ and $y=b$} \\ \frac{6}{2} &\geqslant \cfrac{2}{\cfrac{1}{a} + \cfrac{1}{b}} \qquad \qquad \text{as $a + b = 6$ is given} \\ 3\left(\frac{1}{a} + \frac{1}{b}\right) &\geqslant 2 \qquad \qquad \text{as $a, b > 0$ and so $\frac{1}{a} + \frac{1}{b} > 0$} \\ \frac{1}{a} + \frac{1}{b} &\geqslant \frac{2}{3} \qquad \qquad \text{as required} \end{align*}$

Solution using only AM-GM:

Let $x=a$ and $y=b$ in the statements in the previous post:

$\begin{align*} \frac{x+y}{2} &\geqslant \sqrt{xy} \\ \frac{a+b}{2} &\geqslant \sqrt{ab} \qquad \qquad \text{using $x=a$ and $y=b$} \\ \frac{6}{2} &\geqslant \sqrt{ab} \qquad \qquad \text{as $a + b = 6$ is given} \\ \frac{1}{\sqrt{ab}} &\geqslant \frac{1}{3} \qquad \qquad \text{. . . . . . . . . (1)} \end{align*}$

Now, Let $x=\frac{1}{a}$ and $y=\frac{1}{b}$ in the statements in the previous post (as we can substitute any values that are positive):

$\begin{align*} \frac{x+y}{2} &\geqslant \sqrt{xy} \\ \cfrac{\cfrac{1}{a}+\cfrac{1}{b}}{2} &\geqslant \sqrt{\frac{1}{a} \times \frac{1}{b}} \qquad \qquad \text{using $x=\frac{1}{a}$ and $y=\frac{1}{b}$} \\ \frac{1}{a}+\frac{1}{b} &\geqslant \frac{2}{\sqrt{ab}} \\ &\geqslant 2 \times \frac{1}{3} \qquad \qquad \text{using (1)} \\ \frac{1}{a} + \frac{1}{b} &\geqslant \frac{2}{3} \qquad \qquad \text{as required} \end{align*}$

AbstractBlade · Mar 28, 2021

CM_Tutor said:
Question 25(i)

If $a+b=6$ , and $a, b > 0$ , show that $\frac{1}{a} + \frac{1}{b} \geqslant \frac{2}{3}$

Solution using AM-HM:

Let $x=a$ and $y=b$ in the statements in the previous post:

$\begin{align*} \frac{x+y}{2} &\geqslant \cfrac{2}{\cfrac{1}{x} + \cfrac{1}{y}} \\ \frac{a+b}{2} &\geqslant \cfrac{2}{\cfrac{1}{a} + \cfrac{1}{b}} \qquad \qquad \text{using $x=a$ and $y=b$} \\ \frac{6}{2} &\geqslant \cfrac{2}{\cfrac{1}{a} + \cfrac{1}{b}} \qquad \qquad \text{as $a + b = 6$ is given} \\ 3\left(\frac{1}{a} + \frac{1}{b}\right) &\geqslant 2 \qquad \qquad \text{as $a, b > 0$ and so $\frac{1}{a} + \frac{1}{b} > 0$} \\ \frac{1}{a} + \frac{1}{b} &\geqslant \frac{2}{3} \qquad \qquad \text{as required} \end{align*}$

Solution using only AM-GM:

Let $x=a$ and $y=b$ in the statements in the previous post:

$\begin{align*} \frac{x+y}{2} &\geqslant \sqrt{xy} \\ \frac{a+b}{2} &\geqslant \sqrt{ab} \qquad \qquad \text{using $x=a$ and $y=b$} \\ \frac{6}{2} &\geqslant \sqrt{ab} \qquad \qquad \text{as $a + b = 6$ is given} \\ \frac{1}{\sqrt{ab}} &\geqslant \frac{1}{3} \qquad \qquad \text{. . . . . . . . . (1)} \end{align*}$

Now, Let $x=\frac{1}{a}$ and $y=\frac{1}{b}$ in the statements in the previous post (as we can substitute any values that are positive):

$\begin{align*} \frac{x+y}{2} &\geqslant \sqrt{xy} \\ \cfrac{\cfrac{1}{a}+\cfrac{1}{b}}{2} &\geqslant \sqrt{\frac{1}{a} \times \frac{1}{b}} \qquad \qquad \text{using $x=\frac{1}{a}$ and $y=\frac{1}{b}$} \\ \frac{1}{a}+\frac{1}{b} &\geqslant \frac{2}{\sqrt{ab}} \\ &\geqslant 2 \times \frac{1}{3} \qquad \qquad \text{using (1)} \\ \frac{1}{a} + \frac{1}{b} &\geqslant \frac{2}{3} \qquad \qquad \text{as required} \end{align*}$

does this work for part (ii). I tried it and got something else

CM_Tutor · Mar 28, 2021

Question 25(ii)

If $a+b=c$ , and $a, b > 0$ , show that $\frac{1}{a^2} + \frac{1}{b^2} \geqslant \frac{8}{c^2}$

Solution using only AM-GM:

Let $x=a^2$ and $y=b^2$ in the statements in the previous post:

$\begin{align*} \frac{x+y}{2} &\geqslant \sqrt{xy} \\ \frac{a^2+b^2}{2} &\geqslant \sqrt{a^2b^2} \qquad \qquad \text{using $x=a^2$ and $y=b^2$} \\ \frac{(a+b)^2-2ab}{2} &\geqslant ab \\ \frac{c^2-2ab}{2} &\geqslant ab \qquad \qquad \text{using $a + b = c$, which is given} \\ c^2 &\geqslant 4ab \\ \frac{1}{ab} &\geqslant \frac{4}{c^2} \qquad \qquad \text{. . . . . . . . . (1)} \end{align*}$

Now, Let $x=\frac{1}{a^2}$ and $y=\frac{1}{b^2}$ in the statements in the previous post (as we can substitute any values that are positive):

$\begin{align*} \frac{x+y}{2} &\geqslant \sqrt{xy} \\ \cfrac{\cfrac{1}{a^2}+\cfrac{1}{b^2}}{2} &\geqslant \sqrt{\frac{1}{a^2} \times \frac{1}{b^2}} \qquad \qquad \text{using $x=\frac{1}{a^2}$ and $y=\frac{1}{b^2}$} \\ \frac{1}{a^2}+\frac{1}{b^2} &\geqslant \frac{2}{ab} \\ &\geqslant 2 \times \frac{4}{c^2} \qquad \qquad \text{using (1)} \\ \frac{1}{a^2} + \frac{1}{b^2} &\geqslant \frac{8}{c^2} \qquad \qquad \text{as required} \end{align*}$

Note: I thought than an AM-HM proof would be easier but it turned messy...

Solution using AM-HM:

Let $x=a^2$ and $y=b^2$ in the statements in the previous post:

$\begin{align*} \frac{x+y}{2} &\geqslant \cfrac{2}{\cfrac{1}{x} + \cfrac{1}{y}} \\ \frac{a^2+b^2}{2} &\geqslant \cfrac{2}{\cfrac{1}{a^2} + \cfrac{1}{b^2}} \qquad \qquad \text{using $x=a^2$ and $y=b^2$} \\ (a+b)^2-2ab &\geqslant \cfrac{4}{\cfrac{1}{a^2} + \cfrac{1}{b^2}} \\ c^2-2ab &\geqslant \cfrac{4}{\cfrac{1}{a^2} + \cfrac{1}{b^2}} \qquad \qquad \text{as $a + b = c$ is given} \\ (c^2-2ab)\left(\frac{1}{a^2} + \frac{1}{b^2}\right) &\geqslant 4 \qquad \qquad \text{as $\frac{1}{a^2} + \frac{1}{b^2} > 0$} \\ c^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) - 2ab\left(\frac{1}{a^2} + \frac{1}{b^2}\right) &\geqslant 4 \qquad \qquad \text{. . . . . . . . . (*)} \end{align*}$

Now, we need to simplify:

$2ab\left(\frac{1}{a^2} + \frac{1}{b^2}\right) = 2\left(\frac{ab}{a^2} + \frac{ab}{b^2}\right) = 2\left(\frac{b}{a} + \frac{a}{b}\right)$

and then recognise that the $\frac{a}{b}+\frac{b}{a} \geqslant 2$ , which is easily shown using the AMGM inequality:

$\begin{align*} \frac{x+y}{2} &\geqslant \sqrt{xy} \\ \cfrac{\cfrac{a}{b}+\cfrac{b}{a}}{2} &\geqslant \sqrt{\frac{a}{b} \times \frac{b}{a}} \qquad \qquad \text{using $x=\frac{a}{b}$ and $y=\frac{b}{a}$} \\ \frac{a}{b}+\frac{b}{a} &\geqslant 2 \end{align*}$

So, returning to (*), we have:

$\begin{align*} c^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) - 2ab\left(\frac{1}{a^2} + \frac{1}{b^2}\right) &\geqslant 4 \\ c^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) - 2 \times 2 &\geqslant 4 \\ c^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) &\geqslant 4 + 4 \\ c^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) &\geqslant 8 \\ \frac{1}{a^2} + \frac{1}{b^2} \geqslant \frac{8}{c^2} \end{align*}$

AbstractBlade · Mar 28, 2021

CM_Tutor said:
Question 25(ii)

If $a+b=c$ , and $a, b > 0$ , show that $\frac{1}{a^2} + \frac{1}{b^2} \geqslant \frac{8}{c^2}$

Solution using only AM-GM:

Let $x=a^2$ and $y=b^2$ in the statements in the previous post:

$\begin{align*} \frac{x+y}{2} &\geqslant \sqrt{xy} \\ \frac{a^2+b^2}{2} &\geqslant \sqrt{a^2b^2} \qquad \qquad \text{using $x=a^2$ and $y=b^2$} \\ \frac{(a+b)^2-2ab}{2} &\geqslant ab \\ \frac{c^2-2ab}{2} &\geqslant ab \qquad \qquad \text{using $a + b = c$, which is given} \\ c^2 &\geqslant 4ab \\ \frac{1}{ab} &\geqslant \frac{4}{c^2} \qquad \qquad \text{. . . . . . . . . (1)} \end{align*}$

Now, Let $x=\frac{1}{a^2}$ and $y=\frac{1}{b^2}$ in the statements in the previous post (as we can substitute any values that are positive):

$\begin{align*} \frac{x+y}{2} &\geqslant \sqrt{xy} \\ \cfrac{\cfrac{1}{a^2}+\cfrac{1}{b^2}}{2} &\geqslant \sqrt{\frac{1}{a^2} \times \frac{1}{b^2}} \qquad \qquad \text{using $x=\frac{1}{a^2}$ and $y=\frac{1}{b^2}$} \\ \frac{1}{a^2}+\frac{1}{b^2} &\geqslant \frac{2}{ab} \\ &\geqslant 2 \times \frac{4}{c^2} \qquad \qquad \text{using (1)} \\ \frac{1}{a^2} + \frac{1}{b^2} &\geqslant \frac{8}{c^2} \qquad \qquad \text{as required} \end{align*}$

Note: I thought than an AM-HM proof would be easier but it turned messy...

Solution using AM-HM:

Let $x=a^2$ and $y=b^2$ in the statements in the previous post:

$\begin{align*} \frac{x+y}{2} &\geqslant \cfrac{2}{\cfrac{1}{x} + \cfrac{1}{y}} \\ \frac{a^2+b^2}{2} &\geqslant \cfrac{2}{\cfrac{1}{a^2} + \cfrac{1}{b^2}} \qquad \qquad \text{using $x=a^2$ and $y=b^2$} \\ (a+b)^2-2ab &\geqslant \cfrac{4}{\cfrac{1}{a^2} + \cfrac{1}{b^2}} \\ c^2-2ab &\geqslant \cfrac{4}{\cfrac{1}{a^2} + \cfrac{1}{b^2}} \qquad \qquad \text{as $a + b = c$ is given} \\ (c^2-2ab)\left(\frac{1}{a^2} + \frac{1}{b^2}\right) &\geqslant 4 \qquad \qquad \text{as $\frac{1}{a^2} + \frac{1}{b^2} > 0$} \\ c^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) - 2ab\left(\frac{1}{a^2} + \frac{1}{b^2}\right) &\geqslant 4 \qquad \qquad \text{. . . . . . . . . (*)} \end{align*}$

Now, we need to simplify:

$2ab\left(\frac{1}{a^2} + \frac{1}{b^2}\right) = 2\left(\frac{ab}{a^2} + \frac{ab}{b^2}\right) = 2\left(\frac{b}{a} + \frac{a}{b}\right)$

and then recognise that the $\frac{a}{b}+\frac{b}{a} \geqslant 2$ , which is easily shown using the AMGM inequality:

$\begin{align*} \frac{x+y}{2} &\geqslant \sqrt{xy} \\ \cfrac{\cfrac{a}{b}+\cfrac{b}{a}}{2} &\geqslant \sqrt{\frac{a}{b} \times \frac{b}{a}} \qquad \qquad \text{using $x=\frac{a}{b}$ and $y=\frac{b}{a}$} \\ \frac{a}{b}+\frac{b}{a} &\geqslant 2 \end{align*}$

So, returning to (*), we have:

$\begin{align*} c^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) - 2ab\left(\frac{1}{a^2} + \frac{1}{b^2}\right) &\geqslant 4 \\ c^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) - 2 \times 2 &\geqslant 4 \\ c^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) &\geqslant 4 + 4 \\ c^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) &\geqslant 8 \\ \frac{1}{a^2} + \frac{1}{b^2} \geqslant \frac{8}{c^2} \end{align*}$

thank you

Proofing question (1 Viewer)

AbstractBlade

New Member

Attachments

CM_Tutor

Moderator

CM_Tutor

Moderator

AbstractBlade

New Member

CM_Tutor

Moderator

AbstractBlade

New Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)