Least value of $a$ for which $4ax^2 + \frac{1}{x} \geq 1$

Find the least value of $a \in R$ for which $4ax^2 + \frac{1}{x} \geq 1$, for all $x>0$.

The equation will transform into (Using $x>0$)

$4ax^3-x+1\geq 0$

But I don't know how to deal with this cubic inequality. Could someone give me some hint as how to proceed?

• If $a <0$ then for large $x$ the expression cannot hold. Hence $a \ge 0$. Does the smallest such value work? – copper.hat May 24 '16 at 16:31
• @copper.hat - obviously not, what are you suggesting? – mathguy May 24 '16 at 16:33
• @mathguy: Sorry, I was a little obtuse. I meant to conclude that $a>0$ hence we have a function that has a unique minimiser. – copper.hat May 24 '16 at 16:35

8 Answers

Hint: For $a>0$, we have by AM-GM that $4ax^2+\frac{1}{x}\geq 3\sqrt[3]{a}$. The equality holds iff $x=\frac{1}{2\sqrt[3]{a}}$. For $a<0$, see copperhat's comment. For $a=0$, this is trivial.

If we make a change of variable $a=1/A^3$ and $x=Au/2$, then the problem becomes one of finding the largest value of $A$ for which the inequality

$$u^2+{2\over u}\ge A$$

holds for all $u\gt0$.

If you know calculus, then it's easy to see that $f(u)=u^2+{2\over u}$ is minimized at $u=1$, so that $f(1)=3$ is the largest that $A$ can be. If you don't know calculus, then some extra work is required:

In order for the inequality to hold for all $u\gt0$, we clearly need $A\le1^2+{2\over1}=3$. But it's also easy to see that, for $u\gt0$, we have

\begin{align} u^2+{2\over u}\ge3&\iff u^3-3u+2\ge0\\ &\iff(u-1)^2(u+2)\ge0 \end{align}

Since the final inequality here is obviously true (the product of a square and a positive number is necessarily non-negative), we can conclude that $A=3$ is the largest number for which the inequality $u^2+{2\over u}\ge A$ holds for all $u\gt0$, and thus $a=1/27$ is the smallest value for which the original inequality holds for all $x\gt0$.

As copper.hat has commented, $a>0$. Assuming this is pre-calculus, as tagged, you shouldn't use derivatives to find points of maximum and minimum.

So this leaves elementary math. Can you use the arithmetic-geometric mean inequality theorem?

$$\frac{4ax^2 + \frac 1 {2x} + \frac 1 {2x}}{3} \ge \sqrt[3]{4ax^2 \cdot \frac 1 {2x} \cdot \frac 1 {2x}}$$

and equality is possible when $4ax^2 = \dfrac 1 {2x}$. So $a$ should be so that the RHS is $\dfrac 1 3$. Which leaves $a = \dfrac 1 {27}$.

This method will only give you the least value of $a$ if the cubic has a turning point on $x > 0$, but I think it's a pretty cool method.

Regarding $4ax^3-x+1 \ge 0$:

Consider $y=4ax^3-x+1$

If the graph has a turning point on $x > 0$, then we will want that turning point to be a double root on the $x$ axis. Assume this root is $b$. Then, $y$ will be divisible by $(x-b)^2$. Perform long division until you are about to sum the terms to obtain the remainder. You know that the remainder has to be equal to zero, so you can equate coefficients and solve a system of $2$ equations in $a$ and $b$.

Idea taken from the accepted answer of Finding the range of a $y=-x^2(x+5)(x-3)$ without calculus?

The value of $f(x) = 4ax^2 + 1/x$ approaches infinity as $x\to0$ or $x\to\infty$, so clearly it has a minimum somewhere in between.

The derivative $f'(x) = 8ax - 1/x^2$. Setting this to zero:

$$8ax - 1/x^2 = 0$$ $$8ax^3 = 1$$ $$x_0 = {1\over 2a^{1/3}}$$

Substituting: $f(x_0) = 4 a {1\over 4a^{2/3}} + 2a^{1/3} = (1+2) a^{1/3} = 3 a^{1/3}$.

To find the value where the minimum is 1:

$$f(x_0) = 1 = 3 a^{1/3} \implies a = 1/27$$

Since the multiple of $a$ is $4x^2$ which is always positive, clearly the value increases as $a$ increases, so $a = 1/27$ is the answer. This reaches its minimum at $x = 3/2$.

Let $f(x)=4ax^3-x+1$, then $f'(x)=12ax^2-1$. If $a\le 0$, then $f$ monotonely decreases and violates $f(x)\ge 0$ for all positive $x$. If $a>0$, then $f$ has a critical point at $x=\sqrt{\frac{1}{12a}}$, and $f\left(\sqrt{\frac{1}{12a}}\right)=\frac{1}{3\sqrt{12a}} - \frac{1}{\sqrt{12a}} +1$. It is a minimum value of $f$ on $(0,\infty)$. Can you proceed from this?

• Question is tagged precalculus. – almagest May 24 '16 at 16:41
• @almagest You're right, but there is no context whether OP knows differentiation right now. Also, my post may help other users who can do differentiation, offering a different view of point. – choco_addicted May 24 '16 at 16:57

Use cardanos fromula so find the three solutions $x_1, x_2$ and $x_3$. Then we can rewrite the inequality.

$$4a(x-x_1)(x-x_2)(x-x_3)\geq 0$$

As $4a>0$: $$(x-x_1)(x-x_2)(x-x_3)\geq 0$$

Lets name the linear terms $a,b$ and $c$. If $a,b,c>0$ the inequality is true, you will get three equations solve them for x and find intersection of them. Are there other combinations of the signs so that the product is still positive?

Differentiating the original equation we get $8ax-\frac{1}{x^2}$ so $x={\frac{1}{8a}}^{1/3}$ substituting in original equation we get $a^{1/3}\geq 1/3$ so minimum value of $a=1/{27}$

• tagged as precalculus, that seems to preclude differentiating? – mathguy May 24 '16 at 16:40
• But OP hasn't mentioned anywhere not to use calculus. If he/she posts comment I'll delete my answer – Archis Welankar May 24 '16 at 16:42