Question

The sequence a1,a1,a3 is such that \(a_n>5\) and \(\a_{n+1} = \frac{4a_n}{5}+\frac{5}{a_n}\) for every positive integer n. Prove by mathematical induction that \(a_n>5\) for every positive integer n.

Answer 1

How am I supposed to prove for a(k+1)?

Answer 2

suppose it's true for a(k) then prove it's true for a(k+1)

Answer 3

I think this is a reverse of mathematical induction

Answer 4

I am well aware of the method. I can't however get a conclusion for a(k+1)

Answer 5

I can get to the form
\[a_{k+1}> \frac{25}{a_n} \]
From there, I dont know how to prove \(a_{k+1}>5\)

Answer 6

My method:
\(a_k>5\)
\((a_k)^2>25\)
\(4(a_k)^2>100\)
\(4(a_k)^2+25>125\)
\[\frac{4(a_k)^2+25}{5a_n}>\frac{25}{a_n}\]
\(a_{k+1}>\frac{25}{a_n}\)
Not sure what to do next

Answer 7

@ParthKohli

Answer 8

I found the proof, but in kind of a dirty way. You analyze it as a function
f(x) = (4x^2+25)/5x
You'll find that the function is strictly increasing for x>5 and that the minimum value at x=5 is f(x)=5. Meaning for x>5, f(x)>5
So for a_(k) > 5, a_(k+1)>5

Answer 9

\(\color{#0cbb34}{\text{Originally Posted by}}\) @math&ing001
I found the proof, but in kind of a dirty way. You analyze it as a function
f(x) = (4x^2+25)/5x
You'll find that the function is strictly increasing for x>5 and that the minimum value at x=5 is f(x)=5. Meaning for x>5, f(x)>5
So for a_(k) > 5, a_(k+1)>5
\(\color{#0cbb34}{\text{End of Quote}}\)
Hats off to your brilliance. Thanks

Answer 10

Welcome !

Answer 11

Here is the same solution that I was going to post. I think that I posted for another problem before
\[
g(x)=\frac{4 x}{5}+\frac{5}{x}\\
g'(x)=\frac{4}{5}-\frac{5}{x^2}=\frac{(2 x-5) (2 x+5)}{5 x^2}
\]
so g is increasing as x>2,5 and g(2.5)=5
All the a_n are bigger than at least 4
We are done