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Topic: Regarding the proof that 8^sqrt(7) < 7^sqrt(8)
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J. B. Rainsberger

Posts: 47
Registered: 12/6/04
Regarding the proof that 8^sqrt(7) < 7^sqrt(8)
Posted: Jun 16, 1996 9:31 PM
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I have spent a considerable amount of time attempting to derive the
inequality 8^sqrt(7) < 7^sqrt(8) and have gotten tantalizingly close, but
not completed the thing.

I have seen a few "proofs," but have not understood them in their entirety,

"It suffices to prove that

8^sqrt(7) < 7^[31 sqrt(7) / 29]


8 / 7 < 7^30 / 8^28."

How on Earth do these two inequalities relate? I would appreciate a brief
explanation and/or some references. I do admit that inequalities are a weak
spot where I am concerned. I have no real "feel" for them. I tried to
correct that this year when I developed several of them for a small project
in Real Analysis. Still, I am not comfortable with them.

Wait. . . nevermind. <Grin> I believe it goes like this:

8^sqrt(7) < 7^[31 sqrt(7) / 29]
8^[29 sqrt(7)] < 7^[31 sqrt(7)]
8^29 < 7^31
8/7 < 7^30/8^28

Aha. I wanted to post this anyhow - just in case I was the only one clouded
up about this.

(Ahem) As you were. . . .

J. B. Rainsberger, | Any six-year-old who knows the
York University | words nefarious and extemp-
(collège Glendon) | oraneous gets my vote. | -- Joe on Calvin.

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