# Prove or disprove: For all x sufficiently large, $x^3+3x \geq 7x^2+12$

I'm struggling a bit with coming up with how to prove this. I believe it's true, and selected the value 10 to be my "sufficiently large" x, but I'm not sure sure where to go from here. I initially thought I could prove it by induction, but it doesn't seem like that's very useful.

I could pretty easily prove that the statement is true for the value x=10, but can anyone give me a hint as to how I would go about proving it's true for all x values 10 and above?

• If $x\ge7$, we have $x^3+3x\ge7x^2+21>7x^2+12$. Mar 27 at 4:19
• You can prove it by induction, and if you do it is useful for your purpose. It is more work than Calvin Kohr's approach. You would show that it is true for $n=10$, then show that each increment in $n$ raises the left side more than the right so that the inequality stays true. Mar 27 at 4:26

Try showing $$x^3+3x-(7x^2+12)$$ is increasing for $$x\geq x^*$$.

If $$x>10$$ then $$x^3 > 10x^2 > 7x^2$$ and $$3x > 30 > 12$$ so $$x^3+3x>7x^2+12.$$

I guess you should try this:

$$\begin{cases} x^3≥7x^2 \\ 3x≥12 \end{cases} \Longrightarrow \begin{cases} x≥7 \\ x≥4 \end{cases}$$

$$\implies x≥ \color{red}{{\large{?}}}$$

By the Fundamental Theorem of Algebra, $$P(x)=x^3-7x^2+3x-12$$ has at most 3 real roots. Let $$x^*$$ denote the largest of these. By the Intermediate Value Theorem, $$P(x)$$ must be either always positive or always negative for $$x>x^*$$. From here you only need to show that $$P(x)>0$$ for one $$x>x^*$$ to show that it is true for all $$x>x^*$$. Then, just observe that $$P(x)\ge0\Leftrightarrow x^3+3x\ge7x^2+12$$.

If $$x\ge 7$$ then $$x^3 \ge 7x^2$$

If $$x \ge 4$$ then $$3x \ge 3\cdot 4 = 12$$.

So if $$x \ge \max (4,7)$$ then $$x^3 + 3x \ge 7x^2 + 12$$.

.....

Alternatively. If $$x \ge 0$$ then $$7x^2 + 12 \ge 12 > 0$$ so

If $$x > 0$$ then $$x^3 + 3x \ge 7x^2 + 12 \iff$$

$$\frac {x^2 + 3x}{7x^2 + 12} \ge 1$$.

And if $$x > 0$$ then $$\frac {x^2+3x}{7x^2 + 12} > \frac {x^2 +3x}{7x^2 + 21} = \frac x 7$$. And if $$x \ge 7$$ then $$\frac x7 \ge 1$$.

.....

Or $$(x^3 +3x) \ge 7x + 12 \iff$$

$$(x^3 + 3x) - (7x^2 + 12) \ge 0$$

$$(x^3 + 3x) - (7x^2 + 12)=$$

$$x(x^2 + 3x) - (7x^2 + 12)$$. If $$x \ge 7$$ then

$$x(x^2 + 3x) - (7x^2 + 12)\ge 7(x^2 + 3x) -(7x^2 + 12) = 21x - 12$$

If $$x \ge 1$$ then $$21x - 12 \ge 21 -12 = 9 > 0$$.

So if $$x \ge 7$$ then

$$(x^3+ 3x) - (7x^2 + 12) =$$

$$x(x^2 + 3x) - (7x^2 + 12) \ge$$

$$7(x^2 + 3x) - (7x^2 + 12) =$$

$$21x - 12 \ge$$

$$21 -12 = 9 > 0$$.

So if $$x \ge 7$$ then $$(x^3 +3x) \ge 7x + 12$$

For $$x>0$$ you can divide the inequality by $$x^3$$ and get the equivalent inequality

$$1+\frac 3{x^2} \geq \frac 7x + \frac{12}{x^3}$$

Now before continuing have a careful look at both sides of the inequality:

The left side is obviously always greater than $$1$$. The right side approaches zero for increasing $$x$$. Now, you only need to look for a positive $$x$$, for which the right side is less than or equal to $$1$$.

In your case for $$x\geq 10$$ you get:

$$1+\frac 3{x^2} \geq 1 \geq \frac 7{10}+\frac{12}{1000}\stackrel{x\geq 10}{\geq} \frac 7x + \frac{12}{x^3}$$ This inequality is obviously true and, hence, you are done.