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Prove that, for every natural number $n$, $x^n +n ≥ nx + 1$ for all $x>0$

This is the first part of the question, and I am not too sure how to prove it. My steps to prove it were to find a maximum-minimum case for $x$ and $n$ (ie. when $x → ∞$ and $n → ∞$, but I was not able to find a solution from there. Any help for this is appreciated.

P.S. I also need to prove that this inequality holds for $n= 3/2$, but I think that this can be done if I gain an understanding first of the above proof, so no spoilers please for proving the $n=3/2$ part $:)$

Alan Gardiner
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4 Answers4

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Using Bernoulli's Inequality: $$x^n=(1+(x-1))^n \geqslant 1+n(x-1)=nx+1-n$$

Macavity
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  • is there a way to prove this without Bernoulli's Inequality, because I have not learnt how to use this inequality in class

    also for proving n=3/2 works do I use the bernoulli inequality to expand and simplify?

     – Alan Gardiner Sep 12 '23 at 06:23
  • @AlanG You could re-prove it yourself using induction (as is done in the link in the answer). – Derek Allums Sep 12 '23 at 06:28
  • alright I think I got it - but how does to proof work for $n=3/2$ - is there another inequality I should be aware of @DerekAllums – Alan Gardiner Sep 12 '23 at 06:35
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    (+1) This is how I did this. Bernoulli's Inequality holds for all real $n\ge1$, including $n=3/2$. Bernoulli is proven for integers in this answer, and extend it to rationals in this answer, using induction. – robjohn Sep 12 '23 at 07:45
  • @AlanG As mentioned in comments, Bernoulli's inequality can be proved by induction and extended to rationals. For alternate proofs, you'll need to mention what you are familiar with. For e.g. (for natural $n$), we can directly "see" it as an AM-GM, like so: $x^n+1+1+1\dots+1 \geqslant nx +1$, assuming you're familiar with the same. – Macavity Sep 12 '23 at 09:30
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$$x^n+n-(nx+1)=(x-1)(x^{n-1}+x^{n-2}+\dots+x+1-n)$$ and these two factors are either both $\ge0$ (if $x\ge1$) or both $\le0$ (if $0\le x\le1$).

Anne Bauval
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Define $f(x)=x^n-nx$. Then $f'(x)=nx^{n-1}-n=n(x^{n-1}-1)$. Thus, $f'(x)=0$ only when $x=1$ and $f(1)=1-n$. For $n \geq 2, f''(x)=n(n-1)x^{n-2} \gt 0$, so $x=1$ is an absolute minimum of $f(x)$. If $n=1$, equality (in the original inequality) holds for any $x \gt 0$.

$f(x)=x^n-nx \geq 1-n \Rightarrow x^n+n \geq nx+1$.

But if the precalculus tag is accurate, you may not have access to these methods.

Robert Shore
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I prove the following statement by induction on $n$ (without first having to prove Bernoulli's inequality by induction):

Statement $P(n): \quad x^n +n \geq nx + 1\quad$ for all $x>0.$

Proof of basis case $(n=1): \quad x^1 + 1 \geq x + 1\ $ for all $x>0.\ $ This true statement is $P(1).$ Thus the basis case $P(1)$ is true.

Now suppose $k\in\mathbb{N}$ and suppose $P(k)$ is true, that is,

$$ (1)\qquad x^k + k \geq kx + 1\quad \text{ for all } x>0. $$

We want to show that $P(k+1)$ is true, that is,

$$ x^{k+1} + k+1 \geq (k+1)x + 1\quad \text{ for all } x>0. $$

Let $x>0.$

Since $P(k)$ is true by assumption, multiplying both sides of $(1)$ by $x$ gives:

$$ x^{k+1} + kx \geq k x^2 + x, \implies x^{k+1} \geq kx^2 -kx +x.$$

Adding $k+1$ to both sides gives:

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

$$ = kx^2 -2kx + kx +x + k + 1 $$

$$ = (kx^2 -2kx + k) + kx +x + 1 $$

$$ = k(x-1)^2 + kx +x + 1 $$

$$ \geq kx + x + 1 = (k+1)x + 1. $$

Since $x>0$ was arbitrary, it follows that $x^{k+1} + k + 1 \geq (k+1)x + 1\quad \text{ for all } x>0.\ $ This completes the induction step and therefore the proof also.

Adam Rubinson
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