Proving I_n \leq (2n/(2n+1))^n by induction for positive n

[converting1]

[tex] I_n = \displaystyle \int_0^1 (1-x^2)^ndx, n \geq 0 [/tex]

Given that [itex] (2n + 1)I_n = 2nI_{n-1} [/itex]

proove by induction that

[itex] I_n \leq \left (\dfrac{2n}{2n + 1} \right)^n [/itex] for positive integers of n

in the solutions, could someone explain how they got to step 1, and why we need to show step 2 to complete the proof?

Solutions: http://gyazo.com/26e85134e4d5c13d5d7a49a0de91ae58

 

[milesyoung]

You have the statement:
[tex]
S_n : I_n \leq \left( \frac{2n}{2n + 1} \right)^n
[/tex]
and you want to prove that it's true for all n, where n is a positive integer.
You can do this by proving that [itex]S_1[/itex] and [itex]S_k \Rightarrow S_{k + 1}[/itex] are true, where k is a positive integer.

Proving [itex]S_1[/itex] is true should be simple enough. Prove [itex]S_k \Rightarrow S_{k + 1}[/itex] is true by direct proof, i.e. assume [itex]S_k[/itex] to be true and show that it forces [itex]S_{k + 1}[/itex] to be true:
[tex]
S_k : I_k \leq \left( \frac{2k}{2k + 1} \right)^k \Leftrightarrow \frac{2k + 2}{2k + 3} I_k \leq \frac{2k + 2}{2k + 3} \left( \frac{2k}{2k + 1} \right)^k \Leftrightarrow I_{k + 1} \leq \frac{2k + 2}{2k + 3} \left( \frac{2k}{2k + 1} \right)^k
[/tex]
Here I just multiplied both sides of the inequality by [itex]\frac{2k + 2}{2k + 3}[/itex] and used the relation [itex]I_n = \frac{2n}{2n + 1} I_{n - 1}[/itex].

You have:
[tex]
S_{k + 1} : I_{k + 1} \leq \left( \frac{2(k + 1)}{2(k + 1) + 1} \right)^{k + 1} = \left( \frac{2k + 2}{2k + 3} \right)^{k + 1} = \left( \frac{2k + 2}{2k + 3} \right)^{1} \left( \frac{2k + 2}{2k + 3} \right)^{k}
[/tex]
Thus, if you can show that:
[tex]
\frac{2k + 2}{2k + 3} \left( \frac{2k}{2k + 1} \right)^k \leq \left( \frac{2k + 2}{2k + 3} \right)^{1} \left( \frac{2k + 2}{2k + 3} \right)^{k}
[/tex]
then [itex]S_{k + 1}[/itex] must be true if [itex]S_{k}[/itex] is true.

It follows by induction that [itex]S_n[/itex] is true for all n, where n is a positive integer.

 

[converting1]

You have the statement:
[tex]
S_n : I_n \leq \left( \frac{2n}{2n + 1} \right)^n
[/tex]
and you want to prove that it's true for all n, where n is a positive integer.
You can do this by proving that [itex]S_1[/itex] and [itex]S_k \Rightarrow S_{k + 1}[/itex] are true, where k is a positive integer.

Proving [itex]S_1[/itex] is true should be simple enough. Prove [itex]S_k \Rightarrow S_{k + 1}[/itex] is true by direct proof, i.e. assume [itex]S_k[/itex] to be true and show that it forces [itex]S_{k + 1}[/itex] to be true:
[tex]
S_k : I_k \leq \left( \frac{2k}{2k + 1} \right)^k \Leftrightarrow \frac{2k + 2}{2k + 3} I_k \leq \frac{2k + 2}{2k + 3} \left( \frac{2k}{2k + 1} \right)^k \Leftrightarrow I_{k + 1} \leq \frac{2k + 2}{2k + 3} \left( \frac{2k}{2k + 1} \right)^k
[/tex]
Here I just multiplied both sides of the inequality by [itex]\frac{2k + 2}{2k + 3}[/itex] and used the relation [itex]I_n = \frac{2n}{2n + 1} I_{n - 1}[/itex].

You have:
[tex]
S_{k + 1} : I_{k + 1} \leq \left( \frac{2(k + 1)}{2(k + 1) + 1} \right)^{k + 1} = \left( \frac{2k + 2}{2k + 3} \right)^{k + 1} = \left( \frac{2k + 2}{2k + 3} \right)^{1} \left( \frac{2k + 2}{2k + 3} \right)^{k}
[/tex]
Thus, if you can show that:
[tex]
\frac{2k + 2}{2k + 3} \left( \frac{2k}{2k + 1} \right)^k \leq \left( \frac{2k + 2}{2k + 3} \right)^{1} \left( \frac{2k + 2}{2k + 3} \right)^{k}
[/tex]
then [itex]S_{k + 1}[/itex] must be true if [itex]S_{k}[/itex] is true.

It follows by induction that [itex]S_n[/itex] is true for all n, where n is a positive integer.

managed to show that's true

thank you :)