Inequalities and maximum-minimum of a function

Question

Through my own studies I've often encountered a method for finding function's min/max using standard inequalities. For example:

$a) \ f(x) = {1 \over x} + x \ge 2, x > 0 $, by AM - GM. $b) \ f(x_{1}, x_{2}, \dots, x_{n}) = {({a_{1}x_{1} + \dots + a_{n}x_{n})^{2}} \over x^{2}_{1} + \dots + x^{2}_{n}} \le \sum_{k = 1}^{n}a^{2}_{k}, \ $ by CS.

I cannot find a strict proof for that, and I think it's somehow related to norms. Is there a theorem, that states something like: "Each time function get rid of all variables by an inequality, it attains exact upper/lower bound?"

Answer 1

The answer is NO. $f(x) \leq c$ for all $x$ does not imply that $f$ attains its maximum. Similarly for lower bound. In the first example we also have a point where the bound is attained: $x+\frac 1 x \geq 2$ and $x+\frac 1 x = 2$ when $x=1$. This makes $2$ the maximum value. In your second example the inequality you have stated is actually false. You need a square for the numerator in which case the maximum value is attained when $x_i=a_i$ for all $i$.

Answer 2

If you stated the inequality $$f(x)=\frac{1}{x}+x \ge 1 \text{ for } x \gt 0$$ which you can show by inequalities since at least one of the terms is at least $1$ and the other is at least $0$, this result never achieves the lower bound of $1$.

So you cannot say in general "Each time a function gets rid of all variables by an inequality, it attains exactly the upper/lower bound"