Prove that for $n\geq3$, $n^3\geq3n+5$

Question

Prove that for $n\geq3$,$$n^3\geq3n+5$$

My try:

Let $f(x)=x^3-3x-5$. Clearly $x=1$ and $x=-1$ are critical points where $x=1$ is the local minima so the function $f(x)$ is an increasing function for $x>1$ and the result follows.

Please check and if possible provide another way to solve like induction. Thank you.

It's not enough if it's increasing, you also have to check that it is greater than zero after some time. But that is easy, since $f(3) = 13$, for example. — Sarvesh Ravichandran Iyer, Mar 28 '17 at 12:17

score 3 · Answer 1 · answered Mar 28 '17 at 12:27

3

If $n=k+3$ with $k\ge0$, then

$$n^3=(k+3)^3=k^3+9k^2+27k+27\ge3k+14=3(k+3)+5=3n+5$$

answered Mar 28 '17 at 12:27

Barry Cipra

79,832

score 2 · Accepted Answer · answered Mar 28 '17 at 12:24

Using induction:

$1)$ For $n=3$ we get $3^3\ge 3\cdot 3+5=14$

$2)$ Hypothesis: $n^3\ge 3n+5$

$3)$ For $n+1$

$$(n+1)^3=n^3+(3n^2+3n+1)\ge 3n+5+(3n^2+3n+1)\\ (n+1)^3\ge [3(n+1)+5]+(3n^2+3n-2)\quad (1)$$

but

$$3n^2+3n-2=3n(n+1)-2\ge0\text{ for } n\ge 3$$

then from $(1)$ we get

$$(n+1)^3\ge 3(n+1)+5$$

score 1 · Answer 3 · answered Mar 28 '17 at 12:22

1

As $f'(x)=3(x^2-1)$, so for $x\geq 3$, $f'(x)\geq 0$ hence $f$ is increasing for all $x\geq 3$. Also observe that $f(3)=3^3-3.3-5\geq 0$. Now conclude your answer.

answered Mar 28 '17 at 12:22

Parish

832

score 1 · Answer 4 · answered Mar 28 '17 at 13:43

1

Like this.

$n(n^2-3)\geq3(3^2-3)=18\geq5$

answered Mar 28 '17 at 13:43

Takahiro Waki

2,310

Prove that for $n\geq3$, $n^3\geq3n+5$

4 Answers4