AM GM inequality with constants 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?
$a + 1 \ge 2(a)^{1/2}$
 A: (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 $x\in\Bbb{R}$, $x^2\ge0$.
