Prove that $1 + nx ≤ (1 + x)^n$ , for all real numbers $x > −1$ and integers $n ≥ 2$ by mathematical induction.

Question

My attempt
(1) $n=2$
$x > -1 \Rightarrow (1+2x > -1) \land (x^2 \ge 0) \Leftrightarrow (1+2x) \le (1+x)^2 $
Hence p(2) is true.

(2) We must show that for all integers $k \ge 2, p(k) \Rightarrow p(k+1)$

Suppose for inductive hypothesis, $k \ge 2, p(k)$.
That is $1 + kx ≤ (1 + x)^k$
Then, $1 + kx + x = 1 + (k+1)x ≤ (1 + x)^k + x$

I should get $1 + (k+1)x ≤ (1 + x)^{k +1} $. It seems when I multiply each side by (1+x), the left side gets complex and far away from desired inequlity. On the other hand, when I add 1, it doesn't help since we have to get $(1+x)^{k+1}$ on the right side, $(1 + x)^k + x+1$ doesn't increase the exponent of $(1+x)$.

What should I do next to get out of $1 + kx + x = 1 + (k+1)x ≤ (1 + x)^k + x$?

$1 + kx + x = 1 + (k+1)x ≤ (1 + x)^k$

One thing that may be of interest is that $(1+x)^{k+1}=(1+x)^k+x(1+x)^k$... — abiessu, Feb 20 '16 at 14:23
The case $n=2$ is wrong: $1+2x>-1$ doesn't imply $1+2x\le x^2$. — Bernard, Feb 20 '16 at 14:26
I'm sorry, but $1+2x$ may also be $\ge 0$. Thu inequality is true, but not because $1+2x>-1$. B.t.w. you can start induction at $n=1$. — Bernard, Feb 20 '16 at 14:54
@Bernard There's "x > −1" in the condition of the question. So "x > −1 ⇒ (1+2x > −1)". — buzzee, Feb 20 '16 at 15:40
That is quite true but, I insist, the assertion $1+2x>-1$ is not the reason why $1+2x<(1+x)^2$. — Bernard, Feb 20 '16 at 15:44
@Bernard When you add (1+2x) to each side of $x^2 ≥0$, it becomes $(x+1)^2 ≥ (1+2x)$ — buzzee, Feb 20 '16 at 15:58
@Bernard I don't know what you are trying to say. Do you have another method to show p(2)? — buzzee, Feb 21 '16 at 09:10
Actually, $1+2x\le (1+x)^2$ is true whatever $x$. The hypothesis $x>-1$ is required only for the inductive step. — Bernard, Feb 21 '16 at 10:42

score 1 · Accepted Answer · answered Feb 20 '16 at 14:28

1

HINT: we have to prove $$(1+x)^{k+1}\geq 1+(k+1)x$$ multiplying the inequality $$1+kx\le (1+x)^k$$by $1+x$ we get: $$(1+x)^{k+1}\geq (1+kx)(1+x)=1+x(k+1)+kx^2\geq 1+(1+k)x$$ sind $$kx^2\geq 0$$

answered Feb 20 '16 at 14:28

Dr. Sonnhard Graubner

95,283

Prove that $1 + nx ≤ (1 + x)^n$ , for all real numbers $x > −1$ and integers $n ≥ 2$ by mathematical induction.

1 Answers1