Proof by Induction - Inequalities

AI Thread Summary
The discussion focuses on proving an inequality by induction, with participants providing feedback on the initial approach taken. One user points out that the method used was incorrect, emphasizing the need to start from the assumption that the inequality holds for n=k and to manipulate it towards the desired outcome. Another user clarifies that a specific transformation of the inequality can simplify the proof process, suggesting that fractions can be avoided. The conversation highlights the importance of correctly applying induction principles and considering alternative strategies for tackling the problem. Overall, the thread serves as a collaborative effort to refine the understanding of proof by induction in the context of inequalities.
odolwa99
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Homework Statement



Prove by induction that: (Please see attachment)

Homework Equations



The Attempt at a Solution



Can someone please confirm if I have worked the question out correctly. Many thanks.
 

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Your method is not correct. Start from the inequality you supposed to be true (for n=k) and transform it till you arrive to the desired form. You started from the case n=k+1 and went backwards. That is wrong. ehild
 
It might help you to note that
\frac{1}{(1+r)^n}\le \frac{1}{1+ rn}
is exactly the same as
1+ rn\le (1+ r)^n
so you don't have to worry about the fractions.
 
the second line from the end looks incorrect.
\frac{1}{k\cdot r^2} \le 0 \Leftrightarrow k<0, r\ne 0
why won't you try to attack it from another aspect. for example,
p_1>p2>0 \rightarrow \frac{1}{p_2}>\frac{1}{p_1}
 
Thank you for the help guys. I genuinely appreciate it.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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