Question

Given that x is a positive real number, and n is a positive integer.

Prove the inequality \[^n\sqrt{1+x}-1 \leq \frac{x}{n}\]

Answer 1

Suppose P(n) is the statement \[^n\sqrt{1+x}-1 \leq \frac{x}{n}\]
P(1) is the statement \[(1+x)-1 \leq \frac{x}{1}\]iff\[x \leq x\]which is true.

Answer 2

If all of P(1), ... ,P(k) are true, then in particular P(k) is true.
It must be shown that P(k+1) is true.

P(k) is the statement \[^k\sqrt{1+x}-1 \leq \frac{x}{k}\]

Answer 3

didn't study this one so I am brainstorming with you at the moment...
Maybe it could help that you could write it as...

(1+x)^1/k -1 <= x/k

Answer 4

That actually helps a lot

Answer 5

:( Where did JamesJ go :( he would finish this one in like 2 minutes

Answer 6

all I know is that this question involved logarithms, if it helps

Answer 7

I am not sure. That's what i get
\[(1+x-1)((1+x)^\frac{-1}{n} + (1+x)^\frac{-1-n}{n}... +1)\]

Answer 8

looks like a mean value theorem problem

Answer 9

I'm in bed with my laptop so I can't put the pen on the paper, maybe you could apply logarithms and use the power properties to bring that 1/k "downstairs" ?
As I said, I didn't do it @university so I am just brainstorming :)

Answer 10

apply the MVT to \(f(x)=\sqrt[n]{1+x}\)

Answer 11

Try and figure out what Zarkon is suggesting. Induction isn't the way to go here.

Also agd, look at the detailed feedback I gave you on the last induction proof you wrote. Your statement:
"If all of P(1), ... ,P(k) are true, then in particular P(k) is true.
It must be shown that P(k+1) is true"
is wrong.

Answer 12

Hint: using Z's f(x), consider the difference quotient
\[ \frac{f(x) - f(0)}{x-0} \]