Show that $x^6 - x^4 +2 >0$ for all $x$

Question

Show that $x^6 - x^4 +2 >0$ for all $x$.

I can prove this using calculus and I can prove it by factorising the expression.

Is there any way to prove it another way, perhaps by considering cases of $x$? I need a simpler way to explain to a student that cannot factorise degree $6$ polynomials and use calculus at this stage. They can factorise degree $2$ though, if that helps!

Thanks.

Answer 1

If $x^2 \ge 1 \implies x^6 - x^4 = x^4(x^2-1) \ge 0$, you're done. if $x^2 < 1 \implies x^4 < 1 \implies x^4 < 2 \implies 2-x^4 > 0\implies x^6 -x^4+2 = x^6+(2-x^4) \ge 2-x^4 > 0 \implies x^6 - x^4 + 2 > 0$, you're done also.

Answer 2

Use sum of squares! $x^6>x^4$ for $|x|>1$

Otherwise $|x|<1$ and we have $x^6-x^4+2=(x^3-\frac{1}{2}x)^2-\frac{1}{4}x^2+2>0$ since $2>\frac{1}{4}x^2$

Or, magically, we have

$x^6-x^4+2=(x^3-x)^2+(x^2-1)^2+x^2+1>0$

Answer 3

Setting $y=x^2$ we only need to factor $y^3-y^2+2=(y+1)(y^2-2y+2)$, which is easy, because we see immediately a root, namely $y=-1$.

Answer 4

If $x>1$ or $x<-1$ then $x^6-x^4>0$ and when $-1<x<1 $ then $x^4-x^6<1$

Answer 5

By AM-GM $$x^6+2=2\cdot\left(\frac{1}{2}x^6\right)+2\geq3\sqrt[3]{\frac{1}{4}x^{12}\cdot2}=\frac{3}{\sqrt[3]2}x^4\geq x^4.$$

Answer 6

Note: $$x^6-x^4+2>0 \Rightarrow x^6+2>x^4.$$ Case $1$: $$-1<x<1 \Rightarrow 2>x^4 \Rightarrow x^6+2>x^4.$$ Case $2$: $$x=\pm 1 \Rightarrow 1^6+2>1^4.$$ Case $3$: $$x<-1 \ \ or \ \ x>1 \Rightarrow x^6>x^4 \Rightarrow x^6+2>x^4.$$