Mathematical induction Inequality Question (1 Viewer)

Jayden G · Mar 28, 2022

Hey, I am really stuck with this question. Can anyone pls send me their working out and an explanation on how to do it pls. This would be much appreciated as my exam is only a week away!

ExtremelyBoredUser · Mar 28, 2022

Jayden G said:
Hey, I am really stuck with this question. Can anyone pls send me their working out and an explanation on how to do it pls. This would be much appreciated as my exam is only a week away!

for the 2nd

i)

$(x-y)^2 \geq 0$
$x^2 +y^2 \geq 2xy$ (1)
$x^2 + y^2 + 2xy \geq 4xy$
$(x+y)^2 \geq 4xy$
$(s)^2 \geq 4xy$
$\frac{s}{xy} \geq \frac{4}{s}$ (2)

ii)

making the algebra very clear
$\frac{x^2+y^2}{2} \geq xy$ from (1)
$\frac{x^2+y^2}{2} - xy \geq 0$
$\frac{x^2+y^2}{2} - xy \geq 0$
$x^2 + y^2 -\frac{x^2+y^2}{2} - xy \geq 0$
$x^2 + y^2 -\frac{x^2+y^2 + 2xy}{2} \geq 0$ through factorising the -xy in
$x^2 + y^2 \geq \frac{x^2+y^2 + 2xy}{2}$
$x^2 + y^2 \geq \frac{s^2}{2}$ (3)

Now using inequality from (2)
let x = x^2, y = y^2

$\frac{x^2 + y^2}{x^2y^2} \geq \frac{4}{x^2 + y^2}$
$\frac{4}{x^2 + y^2} \geq \frac{4}{\frac{s^2}{2}}$ from (3)
hence
$\frac{x^2 + y^2}{x^2y^2} \geq \frac{8}{s^2}$
$\frac{1}{y^2} + \frac{1}{x^2} \geq \frac{8}{s^2}$

as req.

Probably a nicer way but this is the one that came to mind

5uckerberg · Mar 29, 2022

First step for the first one

Q for reference $a^{n}+\frac{1}{a^{n}}\geq{a^{n-1}+\frac{1}{a^{n-1}}$ for all positive integers $n\geq{1}$ . Where $a > 0$

First step for n=1
$a+\frac{1}{a}\geq{a^{0}+\frac{1}{a^{0}}}$
$a+\frac{1}{a}\geq{2}$

The assumption step which is $n=k, k\geq{1}$
$a^{k}+\frac{1}{a^{k}}\geq{a^{k-1}+\frac{1}{a^{k-1}}}$
There we are RTP for $n=k+1$
$a^{k+1}+\frac{1}{a^{k+1}}\geq{a^{k}+\frac{1}{a^{k}}$
Using assumption
$a^{k}+\frac{1}{a^{k}}\geq{a^{k-1}+\frac{1}{a^{k-1}}}$ times both sides by a and there we will have
$a^{k+1}+\frac{1}{a^{k-1}}\geq{a^{k}+\frac{1}{a^{k-2}}}$
Add $\frac{1}{a^{k+1}}$ and take $\frac{1}{a^{k-1}}$ on both sides
Performing these steps. We will have.
$a^{k+1}+\frac{1}{a^{k+1}}\geq{a^{k}+\frac{1}{a^{k-2}}-\frac{1}{a^{k-1}}+\frac{1}{a^{k+1}}}$
We want $a^{k+1}+\frac{1}{a^{k+1}}\geq{a^{k}+\frac{1}{a^{k}}$ .
But of course this is where it gets fun
$\frac{1}{a^{k-2}}-\frac{1}{a^{k-1}}+\frac{1}{a^{k+1}}\geq{\frac{1}{a^{k}}}$ . A vital step to prove that $a^{k+1}+\frac{1}{a^{k+1}}\geq{a^{k}+\frac{1}{a^{k}}$ because if that is true then you have finished the induction proof. I imagine this would be the trickiest step.
$a^{2}-a+\frac{1}{a}\geq{1}$ multiplying by $a^{k}$ on both sides because from the Q $a > 0, k\geq{1}$
$a^{2}-a+\frac{1}{a}-1\geq{0}$ .

At this point you go, uh what do I do nor do I have anything to offer.
Well, if in doubt this is when we will use calculus.
Differentiate with respect to a
$2a-1-\frac{1}{a^{2}}=\frac{d}{da}\left(a^{2}-a+\frac{1}{a}-1\right)$
Now you want to find stationary points, $2a-1-\frac{1}{a^{2}}=0$
Multiply by $a^{2}$
It becomes $2a^{3}-a^{2}-1$
$\left(a-1\right)\left(2a^{2}+a+1\right)$
a=1 because the other solution is imaginary
Differentiate that again we will receive
$2+\frac{2}{a^{3}}$ sub in $a=1$ . We will have 4 which clearly is greater than 0 so therefore we have a minimum point. Therefore, we have just proven that $\frac{1}{a^{k}}$ is the minimum value and using $\frac{1}{a^{k}}\leq{\frac{1}{a^{k-2}}-\frac{1}{a^{k-1}}+\frac{1}{a^{k+1}}}$ the indicution proof of
$a^{k+1}+\frac{1}{a^{k+1}}\geq{a^{k}+\frac{1}{a^{k}}$ is complete for $n\geq{1}$

Anounymouse352 · Mar 29, 2022

Jayden G said:
Hey, I am really stuck with this question. Can anyone pls send me their working out and an explanation on how to do it pls. This would be much appreciated as my exam is only a week away!
[/QUOTE]
Hopefully this helps for the 2nd question you sent , i just rearrange the inequailit and said i am going to prove an identity then just subbed in s as x+y, dont show the marker that you rearranged just keep it in your head or scribble it out

Anounymouse352 · Mar 29, 2022

with quetion 1 , i jus came acrss an eddie oo video that explains this question, its 3 videos il send 1 and you would be abe to see the the other 3 from the recomedations

5uckerberg · Mar 30, 2022

IIRC the video you posted here for part 3 Eddie used by cases to prove by mathematical induction for the question but my long post provides an alternative if you want to go into great detail because it uses calculus as it utilises the minimum turning point and the fact that we are proving something is greater than or equal to which justifies the usage of calculus.

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Mathematical induction Inequality Question (1 Viewer)

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