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# Find the smallest real constant π for all integers π β₯ 10.?

Find the smallest real constant π such that 3π 2 + 101π β€ πΓ(2π 2 β 19π + 20) for all integers π β₯ 10.

Update:

Sorry it should be 3π^2 + 101π β€ πΓ(2π^2 β 19π + 20) for all integers π β₯ 10.

### 2 Answers

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- kbLv 72 years agoFavorite Answer
Note that 3n^2 + 101n β€ c(2n^2 β 19n + 20) is equivalent to

(2c - 3) n^2 - (19c + 101)n + 20c β₯ 0.

Letting n = 10 yields

100(2c - 3) - 10(19c + 101) + 20c β₯ 0

<==> 30c β₯ 1310

<==> c β₯ 131/3.

We see that c = 131/3 suffices, by examining the plot (note that the curve is above the horizontal axis for all n β₯ 10):

https://www.wolframalpha.com/input/?i=plot+y+%3D+(...

I hope this helps!

- billrussell42Lv 72 years ago
3π 2 + 101π β€ πΓ(2π 2 β 19π + 20)

??

can you rewrite this in understandable notation?

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