Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have proved this by induction but with direct proof, I'm not really sure where to start. Thank you.

Give a direct proof that, $n^3 > n^2 − 6n + 4$ for all $n ∈ {\mathbb N}$ with $n ≥ 2$ .

share|cite|improve this question
Try completing the cube. – Pedro Tamaroff May 3 '12 at 2:38
There is always induction under the rug.. at least for proving the basic facts used in any proof (e.g. $n^2 \ge 0$) – user2468 May 3 '12 at 2:48
@J.D., there is no induction in my answer, as it is shown for all real numbers $x\geq 1$, and not just for the natural numbers. – Álvaro Lozano-Robledo May 3 '12 at 2:59

5 Answers 5

up vote 5 down vote accepted

Since $n\geq 2$ we can say that $n^3 \geq 2n^2=n^2+n^2\geq n^2+4 >n^2-6n+4$. So $n^3>n^2-6n+4$

share|cite|improve this answer
Do you know where I can find a list of these common inequalities which I should know as a mathematics student. I am referring to $n^{3}\ge2n^{2}$. Regards. – Danny Rancher May 3 '12 at 3:06
That comes from the formula. Since $n\geq 2$ long story short, multiply both sides of the equation by $n^2$ and you have $n^3\geq 2n^2$. the second inequality comes form the fact that $n\geq 2 \implies n^2 \geq 4$ so that's the replacement that happens there. And the final is just because we subtracted something. – Joseph Skelton May 3 '12 at 3:12

$n^3-n^2+6n-4=(n^3-n^2+6n-6)+2=(n-1)(n^2+6)+2>0$ for $n>1$.

share|cite|improve this answer
That was it! --- – Pedro Tamaroff May 3 '12 at 2:39

For $n \geq 1$

$$n^3 \geq n^2$$ $$6n > 4$$

add them together.

share|cite|improve this answer
Very efficient! – André Nicolas May 3 '12 at 3:41

Let $f(x)=x^3-(x^2-6x+4)=x^3-x^2+6x-4$. This is a polynomial, therefore a differentiable function on $\mathbb{R}$. Its derivative is given by $$f'(x)=3x^2-2x+6.$$ The discriminant of the quadratic polynomial is $4-4\cdot 18<0$, thus its roots are imaginary. In particular, since $f'(0)=6$, the derivative $f'(x)>0$ for all $x\in\mathbb{R}$. This implies that $f(x)$ is a strictly increasing function for all $x\in\mathbb{R}$. Since $f(1)=2$, we must have that $f(x)\geq 2>0$ for all $x\geq 1$. In particular, $f(x)>0$ for all $x\geq 1$, and $x^3>x^2-6x+4$ for all $x \geq 1$.

share|cite|improve this answer

For what is worth: $$n^3-n^2+6n-4>0$$



share|cite|improve this answer
There are implied iff between these. – lhf May 3 '12 at 2:58
@lhf Just add them if you want. – Pedro Tamaroff May 3 '12 at 3:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.