I have a question where this is this expression needs to be found where it is the minimum:
$x^2 + \frac{25}{x^2} + 3$ where $x > 0$
There are two options to solving this proof:
(1) Using the Arithmetic-geometric mean inequality
(2) Using the basic fact that $x^2 \geq 0$
Note: Arithmetic-Geometric mean inequality (and its manipulation):
$\sqrt{ab} \leq \frac{a+b}{2} \Rightarrow$
$2\sqrt{ab} \leq a+b$
Option (1):
Take $a = x^2$ and $b = \frac{25}{x^2}$, then:
$2\sqrt{25} \leq x^2+\frac{25}{x^2} \Rightarrow$
$10 \leq x^2+\frac{25}{x^2}$
Now, you say that the Arithmetic-geometric mean inequality is an equality when $a = b$, therefore:
$x^2 = \frac{25}{x^2} \Rightarrow$
$x = \sqrt{5}$
Now, because you know the minimum of the equation $10 \leq x^2+\frac{25}{x^2}$ occurs at $x = \sqrt{5}$, substituting in the minimum $x = \sqrt{5}$ into the equation $10 \leq x^2+\frac{25}{x^2}$ gives you $10$.
And with this information, you now know that the original equation $x^2 + \frac{25}{x^2} + 3$ would just be an additional $+3$, therefore, the minimum of the equation would be $13$. Q.E.D.
Option (2):
Use the fact that:
$\left( x - \frac{5}{x} \right)^2 \geq 0$
now expand, and then:
$x^2 - 10 + \frac{25}{x^2} \geq 0$
$x^2 + \frac{25}{x^2} \geq 10$
Now, add $3$ to both sides to get:
$x^2 + \frac{25}{x^2} +3 \geq 13$.
I was wondering whether the second option proof can actually be concluded at that. It is that proof cannot end with that because the equation might not actually reach that minimum, and because no proof is showed on which number get to that minimum, the proof will be invalid.
Thanks.
You're right: Concluding $f(x)\ge 13$ is not enough to say that $13$ is the minimum value of $f(x)$.
However, one can easily see that there must be some $x$ such that $x-5/x=0$ because this is negative when $x=1$, positive when $x=5$ and continuous between $1$ and $5$.
Then repeating the same calculation as in your argument 2, except with $=$ instead of $\ge$, shows that that particular $x$ actually achieves $f(x)=13$.