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Induction Question (1 Viewer)
- Thread starter js992
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From where you left off, we want to prove this:
^{k+1}\leq (\frac{a^{k+1}+b^{k+1}}{4})+(\frac{ab^k + ba^k}{4}) \leq (\frac{a^{k+1}+b^{k+1}}{2}))
The above would be true if and only if:
\leq (\frac{a^{k+1}+b^{k+1}}{4}) \\ \Leftrightarrow ab^k + ba^k \leq a^{k+1}+b^{k+1} \\ \Leftrightarrow a^{k+1}-ab^k -ba^k +b^{k+1} \geq 0 \\ \Leftrightarrow (a-b)(a^k - b^k)\geq 0)
(a^k-b^k)\geq0 . )
(a^k-b^k)\geq0 .)
Thus, by the exhaustion of cases, we proved the result:
(a^k - b^k) \geq 0 \; for\; a>0\; and \; b>0 )
\leq (\frac{a^{k+1}+b^{k+1}}{4}) )
^{k+1}\leq (\frac{a^{k+1}+b^{k+1}}{2}))
Hence, the result is true for n = k+1 if it is also true for n = k.
By, mathematical induction, the result is true for all positive integers of n.
The above would be true if and only if:
Thus, by the exhaustion of cases, we proved the result:
Hence, the result is true for n = k+1 if it is also true for n = k.
By, mathematical induction, the result is true for all positive integers of n.
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I dont understand the part where you say this is only true if _______From where you left off, we want to prove this:
The above would be true if and only if:
\leq (\frac{a^{k+1}+b^{k+1}}{4}) \\ )
Why does that make it true?
And yeah, i got it from maths online, but their worked solutions are kind of confusing, and they do it by moving everything to one side making it < 0. I was wondering how to do it going from where i had left off.