Using AM-GM inequality to find the minima of a function

Question

Minimize: $$2^{\sin x} + 2^{\cos x}$$

I applied AM-GM inequality to get the result.

It goes like this: $$\frac{2^{\sin x}+2^{\cos x}}{2} \geq \sqrt{2^{\sin x} 2^{\cos x}}$$

Since; $$\sqrt{2^{\sin x + \cos x}} = \sqrt{2^{\sqrt{2} \cos(x-\frac{\pi}{4})}} $$

It becomes; $$2^{\sin x} + 2^{\cos x} \geq 2\sqrt{2^{\sqrt{2} \cos(x-\frac{\pi}{4})}}$$

Minimising RHS we get;

$$2^{\sin x} + 2^{\cos x} \geq 2^{1-\frac{1}{\sqrt{2}}}$$

This happens to be the correct answer but still I don't know how the last inequality proves that $2^{1-\frac{1}{\sqrt{2}}}$ is the minima or the least value in its range. In my opinion, it merely says that there's a number and all outputs of $2^{\sin x} + 2^{\cos x}$ are greater than it.

So my question is:

Can we use AM-GM inequality to find the minima? Will it always work? Or does it simply happen to work in this particular case?

Answer 1

The answer to the question "is there any value of the function less than that number?" is no, since it will violate the AM-GM inequality. So all values of it have to be greater than or equal to the number.

it merely says that there's a number and all outputs of $2^{\sin x}+2^{\cos x}$ are greater than it.

That and there is a value for which it is equal to that number, which is when the terms whose AM and GM we are evaluating are equal. So all outputs of the function is greater than or equal to the number, and strictly cannot be less than it. Isn't that what you exactly mean by a minima in that range?

Answer 2

There is one way to find its minimum value. First of all we should observe that both sin(x) and cos(x) should have negative value to make the expression minimum. Hence we can say that x is in range $[\pi,3\pi/2]$.

By differentiating the expression with respect to x, we get $$ln(2)(2^{sin(x)}cos(x)-2^{cos(x)}sin(x))$$ For minimum value and due to continuity and differentiability of the function,it must be zero. $$ln(2)(2^{sin(x)}cos(x)-2^{cos(x)}sin(x))=0$$ $$2^{sin(x)}cos(x)-2^{cos(x)}sin(x)=0$$ Considering x belongs to $(\pi,3\pi/2)$ $$2^{sin(x)}/{sin(x)}=2^{cos(x)}/cos(x)$$ Since the function $f(x)=2^x/x $ is one-one for range (-1,0),gives$$sin(x)=cos(x)$$ or $$x=5\pi/4$$. We get that for this value of x the double differentiation is positive.Hence concluding to a local minima. To confirm a global minima for $(0,2\pi)$ domain,We can check that for x=$\pi$ and x=$3\pi/2$ the value of expression is greater.