Prove that (a1 + a2 + ... + an)(1/a1 + 1/a2 + ... + 1/an) ≥ n^2 for a1,a2,a...,an ﹥ 0

This video is part of the “Proofs with Mathematical Induction” playlist of my channel

Thanks and enjoy the video!

[craftinators]

I’m a double math and physics major, but all these videos are still so enjoyable and fresh. Never stop doing this ❤

[scruffy-mendeleev]

This was a really really really cool proof. Thanks for sharing, and I’m glad to be watching before your channel blows up.

[Pullen-Paradox]

Nice, and thanks for the lemma. Using a/b + b/a >= 2 you can prove the theorem algebraically. Divide the multiplication up into the sums i = j, i < j, and i >j. Then at the end, substitute f2 - aj/ai in SSi<j(ai/aj) and you will get the result.

[AlexChan-tudw]

I love these videos! Pls keep these going!

[nathanweisman]

It is immediate using Cauchy Schwarz (and then square the inequality)

[instinx]

You can also show that this sum is less than the ratio of the max term and the minimum term times n^2. So, the expression is bounded from below and above which is cool.

This is because the sum across all the numbers has to be less than the maximum times n and the sum across all the reciprocals of the numbers in the set has to be less than the reciprocal of the minimum times n. When multiplying these inequalities together, you get that the sum has to be less than (amax/amin)(n^2) where amax denotes the maximum number in the set and amin denotes the minimum in the set. This only works because all the numbers are positive reals.

[berlinisvictorious]

As someone who doesn't consider induction proof as real proof, this was a slick proof

[zeder.]

This is what im doing in discrete math currently, thanks for sharing.

[hamhead]

Alternatively, you could just apply Cauchy-Schwartz lol

[ddimin]

This inequality is actually a slightly rewritten form of the inequality between arithmetic mean and harmonic mean.

[maryvonnedenis]

Cauchy Schwarz inequality :
u = (sqrt Ai) v = (1/sqrtAi) (u.v)² < IuI² . IvI²

[s.rehman.]

here is another cool way to do it using AM ≥ GM.
∑ai ≥ n x nthroot(∏ai) and ∑1/ai ≥ n x nthroot(∏1/ai)

multiplying both
(∑ai)(∑1/ai) ≥ n^2 x nthroot(∏ai) x nthroot(∏1/ai)

But nthroot(∏ai) x nthroot(∏1/ai) is just 1. So,
(∑ai)(∑1/ai) ≥ n^2

[haldanesghost]

Yes, I’m clapping 😂 does being able to follow but not know where you’re gonna go classify this as “mathematical
Thriller”?

[tranceless]

Sir! Can you prove the rule, where if the sum of an N digit number is divisible by three, then that number is also divisible by three, by induction?