Math Olympiad | Which number is Larger ? | Only for Math Genius!

[@nasrullahhusnan2289]

Note that numerator is 1 less than 1 the corresponding denomerator for each term. Hence each term is 1 minus something which is positibe. Therefore the sum is less than 5.

[@myzoom123]

Applying AM >= GM inequality for each of the fractions in the sum will get the result directly. (Note that the GM is 1)

[@PS-mh8ts]

💯 It's the quickest method.

[@sundareshvenugopal6575]

5 - (1/2022 + 1/2023 .. + 1/2025) + 4/2021 Let x = (1/2022 + ... + 1/2025). Since 1/2021 > 1/2022 > 1/2023 > .. > 1/2025, therefore 4/2021 > x. Hence 5 - x + 4/2021 > 5.

[@Rocio62154]

SENSATIONAL!

[@paulortega5317]

Figure it out in general for x & n S(x,n) = x/(x+1) + (x+1)/(x+2) + .... + (x+k)/(x+k+1) +.... + (x+n-1)/(x+n) + (x+n)/x Expand each x/(x+i) S(x,n) = (1 - 1/(x+1) + 1 - 1/(x+2) + .... + 1 - 1/(x+k+1) + .... + 1 - 1/(x+n) + 1 + n/x Rearrange terms S(x,n) = (n+1) + (n/x - 1/(x+1) - 1(x+2) - ... - 1/(x+k+1) - .... - 1/(x+n) Break down n/x and rearrange terms S(x,n) = (n+1) + (1/x - 1/(x+1)) + (1/x - 1/(x+2)) + .... + (1/x - 1/(x+k+1)) + .... + (1/x - 1/(x+n)) Note 1/x > 1/(x+i) for i = 1 to n ∴ S(x,n) > (n+1) For this problem use x = 2021 and n = 4 S(x,n)/(n+1) converges downward to 1 as x → ∞ I am not sure yet if S(x,n)/(n+1) converges as n → ∞

[@KipIngram]

It's easy to see that the sum is greater than 5. The first term in the sum is less than 1 by 1/2022, the second by 1/2023, the third by 1/2024, and the fourth by 1/2025. But the last term is greater than 1 by 4/2021. We can give one of those 4 to each of the first four terms, and 1/2021 is greater than all of the other fractions, so we gain back more than we lost. You only need seven or eight seconds to see this, without writing a single thing down.

[@vijaymaths5483]

Brilliant 👏

[@bartconnolly6104]

Yes .I noticed the same thing. It would be more of a challenge change the last term to 2230/2226

[@RyanLewis-Johnson-wq6xs]

It’s in my head.

[@vijaymaths5483]

Thanks for your beautiful feedback

[@subratamajumder4675]

5

[@pas6295]

5 is more as the sum of the other is 4.448. only.

[@davidbrisbane7206]

Rewrite as ... 1 - 1/2022 + 1 - 1/2023 + 1 - 1/2024 + 1 - 1/2025 + 1 + 4/2021 Now, since 1/2021 > 1/2022, 1/2021 > 1/2023, 1/2021 > 1/2024, 1/2021 > 1/2025, Then it follows that the quantity on the left is greater than 5.

[@albertoolmos21]

You only prove 4 fractions greater than 1, but not that their sum is greater than 5.

[@prabhushettysangame6601]

Please see the video carefully then you have four fractions added with constant 5( it's only left hand side) afterthen you compare with rightside given constant 5

[@Quest3669]

😂

[@chant9064]

Overly complicated