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 $n ≥ 2$ .
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Try completing the cube. – Pedro May 03 '12 at 02:38
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3There is always induction under the rug.. at least for proving the basic facts used in any proof (e.g. $n^2 \ge 0$) – May 03 '12 at 02:48
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@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 03 '12 at 02:59
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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 $n^3>n^2-6n+4$
Joseph Skelton
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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 03 '12 at 03:06
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3That 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 03 '12 at 03:12
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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 $x \geq 1$.
Álvaro Lozano-Robledo
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For what is worth: $$n^3-n^2+6n-4>0$$
$$n^3-3n^2+3n-1+2n^2+3n-3>0$$
$$(n-1)^3+2n^2+3(n-1)>0$$
Pedro
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