Particle projected from above a dome

[TSny]

I think you have to be careful with that technique. In general, the minimum of the function $f(z) = g(z) + h(z)$ occurs when $g'(z) + h'(z) = 0$ and not when $g(z) = h(z)$. The GM of $\sqrt{g(z)h(z)}$ may be unrelated to the minimum value.

Yes, but in our case the GM is independent of $z$.

In our case, $x = az$ and $y = b/z$ with $a$ and $b$ positive numbers. So, the GM of $x$ and $y$ is GM = $\sqrt{ab}$ for all nonzero values of $z$. Hence the AM of $x$ and $y$ must be $\ge \sqrt{ab}$ for all $z$. $$ \text{AM} = \frac 1 2 (x + y) = \frac 1 2 (az + \frac b z) \ge \sqrt{ab}$$To see that the minimum value of the AM is, in fact, $\sqrt{ab}$, we note that it should be clear that there is a value of $z$ such that $az = b/z$ and, hence, $x = y$. When $x = y$, the AM of $x$ and $y$ equals the GM of $x$ and $y$. So, we can conclude that the minimum value of $\frac 1 2 (az + b/z) = \sqrt{ab}$.

I'm sure that's not the clearest argument. But I hope it makes sense.

 

[PeroK]

Yes, but in our case the GM is independent of $z$.

The technique works as long as the minimum occurs when the two functions are equal. Which is the case here. But, in general, it is not a valid technique for minimizing a function.

 

[TSny]

I agree, it's not a general method of finding a minimum. But, I think it's neat when it applies (which probably isn't that often).

Take $F(x) = \dfrac{8 \cot x}{\sqrt x} + 2 \sqrt x \tan x$

The product of the two terms is the constant 16. The first term monotonically decreases for $0 < x < \pi/2$ and the second term monotonically increases for $0< x < \pi/2$. So, there must be a value of $x$ in this range where the two terms are equal. So, we can use the AM-GM inequality to see that the minimum of $F(x)$ in this range is $2\sqrt{16} = 8$.

 

[PeroK]

I agree, it's not a general method of finding a minimum. But, I think it's neat when it applies (which probably isn't that often).

Take $F(x) = \dfrac{8 \cot x}{\sqrt x} + 2 \sqrt x \tan x$

It applies to any function of the form $af(x) + \frac b {f(x)}$ , which that one is.

 

[TSny]

It applies to any function of the form $af(x) + \frac b {f(x)}$ , which that one is.

Yes.

Edit: I think there is also the condition that the two terms $af(x)$ and $\dfrac b {f(x)}$ be positive.

 

[Steve4Physics]

It applies to any function of the form $af(x) + \frac b {f(x)}$ , which that one is.

Providing $af(x)$ and $\frac b {f(x)}$ are both positive in the domain of interest!

 

[PeroK]

Providing $af(x)$ and $\frac b {f(x)}$ are both positive in the domain of interest!

The main criterion is that their ranges overlap. I forgot to add that above.

 

[TSny]

The main criterion is that their ranges overlap. I forgot to add that above.

Yes, in my example I should have said that the first term decreases from $\infty$ to $0$ while the second term increases from $0$ to $\infty$ for $0<x<\pi/2$. Thus, there is an $x$ in this domain where the two terms are equal.

 

[Bling Fizikst]

In general $(a,b>0)$ , if $z>0$ then $$az+\frac{b}{z}\geq 2\sqrt{az\cdot\frac{b}{z}}=2\sqrt{ab}$$ If $z<0\implies -z>0$ then $$-az-\frac{b}{z}\geq 2\sqrt{-az\cdot\frac{-b}{z}} =2\sqrt{ab}\implies az+\frac{b}{z}\leq -2\sqrt{ab}$$