How can I prove that if $x^5-x^3+x=2$, with $x$ a real number, then $x^6>3$, using only middle school algebra?

Question

How can I prove that if $x^5-x^3+x=2$, with $x$ a real number, then $x^6>3$, using only middle school algebra?

I have been struggling with this one. And I can't find any solution that doesn't rely on derivatives.

Answer 1

Notice the alternating signs in $x^5-x^3+x$, it's tempting to write it as a fraction

$x^5-x^3+x=x(x^4-x^2+1)=x\frac{x^6+1}{x^2+1}$

Now the equation becomes $x\frac{x^6+1}{x^2+1}=2$. To prove $x^6>3$, we just need to prove $\frac{x^2+1}{2}>x$, which is simply $x\neq 1$ and can be easily verified by putting 1 back to the equation.

(As pointed out in the comment, we also need the fact that x>0, which can easily be proved based on the fraction form of the equation.)

Answer 2

$x^5-x^3+x=2\Longrightarrow x^6=2x-x^2+x^4\cdots (1)$

Also we have $\displaystyle x^4=\frac{2}{x}+x^2-1\cdots\cdots \cdots \cdots \cdot (2)$

From $(1)$ and $(2),$ we have

$\displaystyle x^6=2x-x^2+\frac{2}{x}+x^2-1=2\bigg(x+\frac{1}{x}\bigg)-1$

$\displaystyle x^6=2\bigg[\underbrace{\bigg(\sqrt{x}-\frac{1}{\sqrt{x}}\bigg)^2}_{\geq 0}+2\bigg]-1\geq 3$

Equality hold when $x=1$

But from $(1),$ We have $1=2$

So we have $\displaystyle x^6>3$

Answer 3

Here is a more straightforward but much more laborious reasoning that doesn’t use derivatives or any meaningful calculus. First, since

$$2=x^5-x^3+x=x\cdot\underbrace{(x^4-x^2+1)}_{>0},$$

we have that $x$ is positive. Now we prove that $f(x)=x^5-x^3+x$ strictly increases. Let $0<x<y$. We want to prove that

$$x^5-x^3+x < y^5-y^3+y$$

$$(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4-x^2-xy-y^2+1)<0.$$

Since $x-y<0$, we want to prove that the second factor is positive. It follows from these AM-GMs:

$$x^4+\frac13\ge 2\sqrt{x^4\cdot\frac13}\ge x^2,$$

$$y^4+\frac13\ge 2\sqrt{y^4\cdot\frac13}\ge y^2,$$

$$x^2y^2+\frac13\ge 2\sqrt{x^2y^2\cdot\frac13}\ge xy.$$

Now, knowing that $f(x)$ is strictly increasing, let us see that $x>\frac65$. Indeed, otherwise $$x^5-x^3+x<\left(\frac65\right)^5-\left(\frac65\right)^3+ \frac65=\frac{6126}{3125}<2.$$

Now $$x\cdot(x^5-x^3+x)=x\cdot 2$$

$$x^6-x^4+x^2=2x$$

$$x^6=x^4-x^2+2x.$$

Knowing that, we consider

$$x^6>3$$

$$x^4-x^2+2x-2>1$$

$$x^2(x+1)(x-1)+2(x-1)>1$$

$$(x-1)(x^2(x+1)+2)>1$$

Since $x>\frac65$ the LHS is a product of two positive increasing functions and thus is increasing. So if this inequality is true for $\frac65$ then it’s true for $x$:

$$\left(\frac65\right)^4-\left(\frac65\right)^2+2\cdot\frac65=\frac{1896}{625}>3.$$

Proof complete.

Answer 4

Here's an answer I feel that would be about the level of middle school (but coming from a high schooler)... I've only used a tiny bit of basic AM-GM Inequality:

$x^5-x^3+x=2$. Taking out x as a common factor and since x is clearly non zero, we get:

$x^4-x^2+1=\frac{2}{x}$. I will use this later.

Now in our original equation, we can rewrite as: $x^5=x^3-x+2$ Multiply x on both sides here to obtain

$x^6=x^4-x^2+2x$. Use from earlier that $x^4-x^2+1=\frac{2}{x}$ to obtain

$x^6=\frac{2}{x}-1+2x$. Using AM-GM Inequality we get $\frac{2}{x}+2x$ >= $2\sqrt{(2x)(\frac{2}{x})}$.

Thus $x^6>= 4-1$ (Equality holding at $2x=\frac{2}{x}$ or $x=1,-1$). Now notice that when we put $x=1$ into $x^5-x^3+1$, we get $1$ and not $2$. Neither does putting x=-1 satisfy the original equation. Which would imply $x^6>3$, equality removed.