3

I was trying to find the minimum value of $a + \frac{1}{a} + 3$ given that $a$ is a positive real number. I found $2$ ways to do it.

Minimum value of $a + \frac{1}{a}$ is $2$ using the AM-GM inequality and therefore the minimum value of $a + \frac{1}{a} + 3$ is $2 + 3 = 5$

Then, I tried using AM-GM inequality directly for the given expression and I got $a + \frac{1}{a} + 3 \ge 3\cdot(3)^{1/3}$

Can anyone please tell me what is the error using the second method?

And why is this true? $a + 1 \ge 2(a)^{1/2}$

user729424
  • 5,061
thornsword
  • 687

1 Answers1

3

(1) You correctly used the AM-GM inequality when you showed that

$$a+\frac{1}{a}\ge2,$$

and of course, this implies that

$$a+\frac{1}{a}+3\ge5.$$

Also, you correctly used the AM-GM inequality when you showed that

$$a+\frac{1}{a}+3\ge3\cdot3^{\frac{1}{3}}.$$

Note that

$$5>3\cdot3^{\frac{1}{3}}\approx4.32,$$

so your first use of the AM-GM inequality yielded a stronger result than your second use did.

You can also note that your first result

$$a+\frac{1}{a}+3\ge5,$$

is the best possible, since $a+\frac{1}{a}+3=5$ if $a=1$.

(2) You can prove that for any $a\ge0$ that $a+1\ge2\sqrt{a}$ by using the AM-GM inequality. Note that for $a\ge0$

$$a+1\ge2\sqrt{a}\quad\text{iff}\quad\frac{a+1}{2}\ge\sqrt{a}$$

and the inequality on the right follows from using AM-GM inequality on $1$ and $a$.

A second way to prove that $a+1\ge2\sqrt{a}$ for any $a\ge0$, is as follows. Note that the following are equivalent:

$$a+1\ge2\sqrt{a}$$ $$a-2\sqrt{a}+1\ge0$$ $$\left(\sqrt{a}-1\right)^2\ge0$$

And the last inequality holds, since for any $x\in\Bbb{R}$, $x^2\ge0$.

user729424
  • 5,061
  • If a + 1/a + 3 >= 3(3)^1/3 , then we must have that the equality exists if a = 1/a = 3 which is not possible right? What's wrong here? The inequality suggests that equality must hold when all the terms are equal.. – thornsword Feb 19 '20 at 03:59
  • In general $a\ge b$ means that either $a>b$ or $a=b$. So if you know that $a>b$, you automatically get that $a\ge b$. We know that $a+\frac{1}{a}+3>3\cdot3^{1/3}$ since $a+\frac{1}{a}+3\ge5$ and $5>3\cdot3^{1/3}$. And since we know that $a+\frac{1}{a}+3>3\cdot3^{1/3}$, it follows that $a+\frac{1}{a}+3\ge3\cdot3^{1/3}$. – user729424 Feb 19 '20 at 04:09
  • 1
    I think what you're confused about is the condition for equality in the AM-GM inequality. Note that AM-GM says two things: (1) $\frac{1}{n}\sum_{i=1}^n a_i\ge\left(\prod_{i=1}^n a_i\right)^{1/n}$ and (2) $\frac{1}{n}\sum_{i=1}^n a_i=\left(\prod_{i=1}^n a_i\right)^{1/n}$ iff $a_1=a_2=\ldots=a_n$. Note that AM-GM inequality gives us that $a+\frac{1}{a}+3\ge3\cdot3^{1/3}$. It also gives us that $a+\frac{1}{a}+3>3\cdot3^{1/3}$ because we would have $a+\frac{1}{a}+3=3\cdot3^{1/3}$ iff $a=\frac{1}{a}=3$, which of course is impossible. – user729424 Feb 19 '20 at 04:10
  • Oh okay, I get it now. Thanks for you help!! – thornsword Feb 19 '20 at 04:16
  • Awesome! Thanks for posting your question! – user729424 Feb 19 '20 at 04:17