# Proving that if $x^4 + 5x + 1 < 27$ then $x < 2$

I need help trying to figure out how I can prove a statement like this. So looking at this I can conclude that this statement is of the form $$P \Rightarrow Q$$

$$P: x^4 + 5x + 1 < 27$$ $$Q: x < 2$$

I wanted to try and prove this by contrapositive , so this state would become

If $$X \geq 2$$ then $$x^4 + 5x + 1 \geq 27$$

Over here I was not sure if I could plug in the value 2 or anythin greater than 2 to see if this is true. plugging in 2 gets me $$2^4 + 5 \times 2 + 1 = 27$$ and since $$27 \geq 27$$ this statement is True.

Am I allowed to prove it like this? Is there a different way to prove a question like this?

• No, you have to prove for a general $x\geq 2$. It might work for a specific value but not in general. That is the problem with your approach – Heisenberg Oct 20 '18 at 22:45
• If $x \ge 2$ then $x^4 + 5x +1 \ge 16+10+1 = 27$. – copper.hat Oct 20 '18 at 22:47
• You can use that for $c\ge0$, $a\ge b$ implies $ac\ge bc$. So for $x\ge 2$ (and thereby $x>0$) we have $x^2\ge 2x\ge 4$, $x^3\ge 2x^2\ge 8$, $x^4\ge 2x^3\ge 16$, and hence by addimg the inequality $x^4+4x+1\ge 16+10+1=27$ indeed – Hagen von Eitzen Oct 20 '18 at 22:48

Suppose $$x\geq2$$. Then $$x^4\geq16\implies x^4 +5x\geq16+5x\geq 16+...$$. Got the hint?

• Are you trying to make the left side look like what we need to prove? – Rufyi Oct 20 '18 at 23:01
• If I start from x^4 >= 16, I would get x^4 + 5x + 1 >= 16 + 5x + 1, can i plug in 2 in for 5x on the right side? – Rufyi Oct 20 '18 at 23:05
• Yes exactly!... – Heisenberg Oct 20 '18 at 23:05
• Can I do what I said in the second comment? – Rufyi Oct 20 '18 at 23:09
• Yes when you plug in 2, you get 27 on the right side thus proving $x^4+5x+1\geq 27$ – Heisenberg Oct 20 '18 at 23:11

A marginally different take:

Let $$p(x) = x^4+5x-26$$, we would like to show that if $$p(x)<0$$ then $$x <2$$.

Note that $$p(2) = 0$$ and so synthetic division gives $$p(x) = (x-2)(x^3+2x^2+4x+13)$$.

In particular, note that $$x^3+2x^2+4x+13$$ has no positive roots. Hence if $$x \ge 0$$, we see that $$p(x)$$ and $$x-2$$ have the same sign, hence if $$p(x) < 0$$ we must have $$x < 2$$.

Trying to show the contrapositive is a valid method, but you aren't allowed to just give an example. For example, if you wanted to prove that

$$x\geq 2\implies x^4+5x+1\leq 27,$$

which is false for example at $$x=0$$, you can't just give the example of $$x=2$$.

One hint for trying to prove this is to show that $$f(x)=x^4+5x+1$$ is increasing over the positive real numbers - in other words, assume that $$0 and show that $$f(b)>f(a)$$.

If $$x\ge 2$$ is it true that $$x^4 \ge 16$$?

If $$x \ge 2$$ is it true that $$5x \ge 10$$?

So if $$x \ge 2$$ is it true that $$x^4 + 5x + 1 \ge 16 + 10 + 1 = 27$$?

...

It could get tedious to go to axioms and prove that if $$x\ge 2 > 0$$ then $$x^4 \ge 2x^3 \ge 4x^2 \ge 8x \ge 16$$ via the axiom: if $$a > 0$$ and $$m < n$$ then $$am < an$$ (applied three times); and $$5 > 0$$ so $$x\ge 2 \implies 5x \ge 2*5 =10$$ by the same axiom; and by the axiom $$a + c > b + c \iff a > b$$ then $$x^4 + 5x + 1 \ge x^4 + 10 + 1 \ge 16+ 10 + 1=27$$. But I think it is safe to assume if we had to prove those we could.

This may be tedious over kill but:

If $$x \ge 2$$ then let $$d = x - 2 \ge 0$$. Then $$x = 2 + d$$

Now because $$d \ge 0$$ then $$d^k \ge 0$$.

So $$x^4 = (2 + d)^4 = 2^4 + 4*2^3*d + 6*2^2*d^2 + 4*2*d^3 + d^4 \ge 2^4$$

And $$5x = 5*(2+d) = 5*2 + 5*d \ge 5*2$$

So $$x^4 + 5x + 1\ge 2^4 + 5*2 + 1 = 27$$.

That should do it.....