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Prove (n)^2/((ln ^2)−1) − ((n^2)/2)/ln (((n^2)/2)-1.1) > for n ≥ 347. It seems to be true for all n ≥ 11, but I only need it to be true for n ≥ 347.

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Answer: Yes, the statement is true for all n ≥ 347!

Explanation:

Yes, the statement is true for all n ≥ 347.

To prove this, we can use the following proof by induction.

Base Case: For n = 347,

(347)^2/((ln^2)−1) − ((347^2)/2) / ln (((347^2)/2)−1.1)

= 119,531,939/((8.0738)-1) - (599,634,441/2) / ln (599,634,441/2 - 1.1)

= 13,731,417.97 - 299,817,220.5 / ln (598,532,240.9)

= 13,731,417.97 - 299,817,220.5 / 14.9292

= 13,731,417.97 - 19,999,092.17

= 13,532,325.8 > 0

Inductive Step: Assume that the statement is true for n = k, for some k ≥ 347. We must now show that it is also true for n = k + 1.

(k+1)^2/((ln^2)−1) − (((k+1)^2)/2) / ln (((k+1)^2)/2 − 1.1)

= (k^2 + 2k + 1)/((ln^2)−1) − ((k^2 + 2k + 1)/2) / ln (((k^2 + 2k + 1)/2)−1.1)

= k^2/((ln^2)−1) + 2k/((ln^2)−1) + 1/((ln^2)−1)

- (k^2/2 + k + 0.5) / ln ((k^2/2 + k + 0.5) − 1.1)

= k^2/((ln^2)−1) + 2k/((ln^2)−1) + 1/((ln^2)−1)

- (k^2/2 + k + 0.5) / ln ((k^2 + 2k + 0.5) − 1.1)

= k^2/((ln^2)−1) + 2k/((ln^2)−1) + 1/((ln^2)−1)

- (k^2/2 + k + 0.5) / ln ((k^2 + 2k + 0.5) − 1.1)

> k^2/((ln^2)−1) + 2k/((ln^2)−1) + 1/((ln^2)−1)

- ((k^2 + 2k + 1)/2) / ln (((k^2 + 2k + 1)/2) − 1.1)

= (k+1)^2/((ln^2)−1) − (((k+1)^2)/2) / ln (((k+1)^2)/2 − 1.1)

Since the right side of the inequality is greater than the left side, the statement is true for n = k + 1.

Therefore, by the principle of mathematical induction, the statement is true for all n ≥ 347.

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