Pls help (1 Viewer)

5uckerberg · Mar 6, 2022

I think we have seen this Q being asked many times on Bored of Studies. I recall this is the third time.

Anyways, start with this

1. $\left(\sqrt{a}-\sqrt{b}\right)^{2}>0$ because $a, b > 0$ .
Next,

2. $\left(\sqrt{c}-\sqrt{d}\right)^{2}>0$ because $c, d > 0$ .

Expanding 1. and 2. we will have respectively

$a+b > 2\sqrt{ab}$
$c+d > 2\sqrt{cd}$

Dividing by 2 we will have
$\frac{a+b}{2} > \sqrt{ab}$
$\frac{c+d}{2} > \sqrt{cd}$

Now, notice that

$\frac{\frac{a+b}{2}+\frac{c+d}{2}}{2} > \sqrt{\sqrt{ab}\sqrt{cd}}$
$\sqrt{ab}\sqrt{cd}\geq{\sqrt{abcd}}$
$\frac{\frac{a+b}{2}+\frac{c+d}{2}}{2} > \sqrt[4]{abcd}$
$\frac{a+b+c+d}{4} > \sqrt[4]{abcd}$

=)(= · Mar 6, 2022

My advice would be to burn that 4u maths in focus text book, it is the most error filled 'maths guide' to ever exist used fitzpatrick instead

yanujw · Mar 8, 2022

5uckerberg said:
Dividing by 2 we will have

Now, notice that

@5uckerberg can you expand on the link between these two steps? Is it adding/multiplying the inequalities or performing AM-GM on them?

Octavius · Mar 8, 2022

yanujw said:
@5uckerberg can you expand on the link between these two steps? Is it adding/multiplying the inequalities or performing AM-GM on them?

You use the AM/GM inequality again

5uckerberg · Mar 9, 2022

yanujw said:
@5uckerberg can you expand on the link between these two steps? Is it adding/multiplying the inequalities or performing AM-GM on them?

AM-GM inequality as @Octavius said.

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Pls help (1 Viewer)

Octavius

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5uckerberg

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=)(=

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yanujw

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Octavius

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