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[lofty cobalt]

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[normal monolith]

Substituting p = q

We easily get √pq

But i need a legitimate method

[ Question 37 ]

Let me know if this method is right

( p^ 1/n + q^ 1/n )/2 ≥ √ (pq)^1/n

At equality, p ^ 1/n = q^ 1/n

But p ≠ q

Hence n = ∞

So we can substitute ( p^ 1/n + q^ 1/n )/2 = √ (pq)^1/n

And hence answer is √pq

[limber timber]

Why considering equality?

[normal monolith]

n approaching infinity

Makes p^1/n = q^1/n

Hence AMGM equality holds

[limber timber]

Well Alright tho I would do it using taylor series of e^x

[normal monolith]

e^x???

[limber timber]

That's a long method but works

[normal monolith]

Not a^x ?

[limber timber]

p^1/n = e^(1/n(lnp))

[normal monolith]

Oh

Yeah

[limber timber]

You cam convert it by common sense don't memorise

[normal monolith]

But

Its easy

[limber timber]

Yeah I just found another way

[normal monolith]

Anyways.ill.try the expansion thkng

I wrote the series wrong om my.first attempt

[limber timber]

I gtg for now

[normal monolith]

3 methods successful 😤😤

[normal monolith]

+close

[formal forge]

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