Mathematical proof that 2020 is the worst!

A light-hearted video containing a few tricks to prove inequalitites.

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[sinecurve]

2017 is prime.
2018 is 2 times a prime.
2019 is 3 times a prime.
2020 is COMBO BREAKER!!!

[Bodyknock]

The lemma “bigger is isomorphic to better” is left as an exercise for the viewer. 😉

[afm]

How large is (2021/2020)^2020? Well, this is (1+1/2020)^2020 which is close to Euler's number e=2.71828...

[goodplacetostop]

14:08 2020 wasn’t that bad because it’s the year I’ve discovered your channel ♥️

[xCorvusx]

Isn't the second inequality trivial?
2021^(1/sqrt(2019)) is bigger than 2020^(1/sqrt(2020)) because 2021 is bigger than 2020, and sqrt(2019) is smaller than sqrt(2020): i. e. the reciprocal of the former is bigger than the reciprocal of the latter.
So all things considered, on the right side you have a bigger number as the base and a bigger number as the exponent.

[routemath]

Man these problems are really neat. I don't think I'd even have a clue where to start many of them before watching your channel. You've opened me up to so many new problem-solving techniques in 2020/2021.

[cartermurphy]

5:50
“...which is larger than 1^2020, which is larger than 1”

[SlidellRobotics]

First problem becomes trivial when you realize that each side is the geometric mean of a set of numbers, and the larger set is the smaller set unioned with a number larger than any in the smaller set. The second is obvious when you realize that 1/√2020 < 1/√2019 as well as 2020 < 2021.

[leif_p]

Finally, a formal proof of something a lot of people have independently conjectured.

[SinclairLocke]

1^2020 is definitely not larger than 1 :p

[andrewparker]

I took the lazy-log way out on this. For both problems, taking logs leads to a pretty neat solution. The first one is a little more lengthy, but the second is 4 lines. Just assume 2020^(1/sqrt(2020)) <= 2021^(1/sqrt(2019) then, take logs:

<=>
sqrt(2019)*ln2020 <= sqrt(2020)*ln2021
<=>
2019(ln2020)^2 <=2020(ln2021)^2
<=>
2019((ln2020)^2 - ln(2021)^2) <= ln(2021)^2

which is obviously true as the LHS is negative but the RHS is positive

[ncantor]

An interesting way to solve the first problem:
By taking both sides to the 2021st power and dividing by 2020!, we can see that it is equivalent to showing that,
2021 > ²⁰²⁰√(2020!)
By the AM-GM inequality,
(1+2+3+...+2020)/2020 >= ²⁰²⁰√(2020!)
The numerator of the LHS is a triangular number, which we can calculate easily
LHS = 1/2*(2020)(2021)/2020 = 2021/2
And, clearly 2021 > 2021/2 >= ²⁰²⁰√(2020!)
which is what we were trying to show.

[przemezio]

11:42 g'(x) should be (1-x)/(x+1)^2

[BigDBrian]

the second question is trivial. Larger base, smaller exponent -> larger value

[darreljones]

The first part can be extended for any pair of numbers to show that the function f(n) defined as the nth root of n factorial is an increasing function, i.e., the value always increases as n increases.

[boristerbeek]

Hi Michael, really nice video as always! However, it might make things a bit clearer if you explicitly write out discussions such as the one you started at the 4:20 mark, to avoid confusion. I was able to follow it, but I feel it may confuse other viewers.

[mysteriousstranger]

Here’s an interesting question:
What are the only values of A and B on the interval [0, 9] such that the three-digit number “A0B” divided by the two-digit number “AB” results in an integer?

As a follow-up question, what happens when A and B can be any positive integer (for example, if A was 7 and B was 13, the result would be 713/(70+13), or 713/83).

[Lucas-mzld]

The first one could be solved as the following:
Let m=(2020!)^(1/2020)
Then m^2020 = 2020!
It's easy to see that 2021 > m, because m is the geometric average of numbers < 2021.
Now (2021!)^(1/2021) = (2021 * m^2020)^(1/2021) > (m * m^2020)^(1/2021) = m, and that's what we wanted

[DubioserKerl]

All I'm hearing is "Tweny Tweny Tweny Tweny..."

[tomatrix]

At 4:31 the answer is quite obvious. The top is larger because 2021 times itself 2020 times is larger thwn 2021! Which consists of only 2020 terms decreasing constantly. Clearly the fraction value is greater than one therefore....