Is this proof of Bernoulli’s inequality correct?

Question

I have to prove $(1+x)^{n} >1+nx$ for $n=2,3,4.... $and $x>-1$ and $x$ isnt 0. There are a lot of proofs of this but l want to know if this one works. If not, can u show where my reasoning is weak.

If $x>-1$ then $1+x>0$ Hence $ (1+x)^{n}>0$ for $n>=2$ and $x>-1$ . For $1+nx$ . I can show that $1+nx=0$ for $n=2,x=-.5$

Since $nx+1>0$ isn't true for a single pair (n,x) in the list of all pairs, the statement that it is true for all pairs of (n,x) is also false.

So $(1+x)^{n}>1+nx$ for $n=2,3,4.... $and $x>-1$.

@chrystomath. $ (1+x)^{n}>0$ for all n>=2 but 1+nx<0 for a single pair ,then it can't be true for all pairs . We don't need to verify further on. That's how l am reasoning, can u explain how this reasoning produces false results. — Aristotle Stagiritis, Aug 04 '20 at 16:16
Looks like a complete non sequitur to me. Can you at least prove that $(1+x)^2\ge1+2x$? — Angina Seng, Aug 04 '20 at 16:20
@angina sure 1+2x+x^2 >=1+2x, equal when x=0. Wait x cant be zero in bernoulli inequality. : ) — Aristotle Stagiritis, Aug 04 '20 at 16:26
I'm also not sure I understand where your conclusion came from. Could you maybe edit the question to explain in a little more details for us? — Riemann'sPointyNose, Aug 04 '20 at 17:13
@Riemann'sPointyNose I give a counter example to the statement nx+1>0 for n>=2 and x>-1, where x isn't zero. A single counter example should suffice to prove it wrong. Since l have proved that$ (1+x)^{n}>0$ for the conditions,$ (1+x)^{n}$> nx+1. — Aristotle Stagiritis, Aug 04 '20 at 17:26
Why would a single counterexample work? Proving there exists a ${(n,x)}$ pair such that ${1 + nx\leq0}$ does not prove for every ${(n,x)}$ pair that ${(1+x)^n>1+nx}$. How do you know there isn't some ${(n,x)}$ pair such that ${1+nx>0}$ and ${(1+x)^n > 1+nx}$? — Riemann'sPointyNose, Aug 04 '20 at 17:31
@Riemann'sPointyNose I think l sort of get your point. If $1+nx>0$ , we cannot say anything about its relation to $(1+x)^{n}$ . So if l prove $1+nx<0 $ for a certain $(n,x)$ , l cannot claim that it is less than $(1+x)^{n}$ for all $(n,x)$ . Correct or not ? — Aristotle Stagiritis, Aug 04 '20 at 17:40
Correct. You can see my answer for an analogy and also another method of proof — Riemann'sPointyNose, Aug 04 '20 at 17:47
One last thing to note - it should be ${\geq}$, not ${>}$. If ${x=0}$, both sides are equal to eachother. It was probably a typo but I thought I'd just point it out incase — Riemann'sPointyNose, Aug 04 '20 at 17:55

Riemann'sPointyNose · Accepted Answer · 2020-08-04T18:03:32.627

Note that your proof is not valid, since we are not trying to disprove anything - so a counterexample approach won't work. "It's not true for a single pair, so it's not true for all pairs" is the incorrect reasoning. That's like saying "every car I've seen today has been red, so every car is red". It doesn't work.

Instead, you could prove the statement by induction. Take arbitrary ${x>-1}$ and ${n \in \{2,3,4,5,6...\}}$. Then for ${n=2}$, the statement is clear

$${(1+x)^2 = 1 + 2x + x^2 \geq 1 + 2x}$$

(since ${x^2\geq 0}$ for any ${x}$). Hence the statement ${(1+x)^n \geq 1+nx}$ is true for ${n=2}$. Now assume that for some ${n=k}$ where ${k \in \{2,3,4,...\}}$ that

$${(1+x)^k \geq 1+kx}$$

Then notice

$${(1+x)^{k+1}=(1+x)(1+x)^k\geq (1+x)(1+kx)=1+kx+x+kx^2 = 1+(k+1)x + kx^2}$$

And again since ${kx^2}$ will be ${\geq 0}$ we have

$${1+(k+1)x + kx^2 \geq 1+(k+1)x}$$

So

$${(1+x)^{k+1}\geq 1+(k+1)x}$$

Since we have shown that the statement is true for ${n=k+1}$ if it's true for ${n=k}$, and we have shown it's true for ${n=2}$ we have the statement is true also for ${n=3,4,5....}$ and so by the Principle of Mathematical Induction we now have that

$${(1+x)^{n}\geq 1+nx}$$

for ${n \in \{2,3,4,5,6,...\}}$ and ${x>-1}$, as required.

Is this proof of Bernoulli’s inequality correct?

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