<s>[INST] Find the global minimum of a function of two variables without derivatives If $f:(\mathbb{R}^{+*})^2\to\mathbb{R}$ is defined by $$f(x,y)=\sqrt{x+y}\cdot\Big(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\Big),$$ 
how would you prove that $f$ has a global minimum without using all the differential calculus tools ? I am asked to only use the inequality $x+y\geq 2\sqrt{xy},$ but steps as $f(x,y)\geq 2\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+y}}\geq 2$ are too violent and do not seem to give a reachable minimum. Thank you for your help ! 
 [/INST] $$f(x,y) = \sqrt{x+y} \cdot \left( \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} \right) =  \sqrt{x+y} \frac{\sqrt{x} + \sqrt{y}}{\sqrt{xy}}$$
Squaring we get 
$$f(x,y)^2 = (x+y) \frac{x + y + 2 \sqrt{xy}}{xy} = \frac{(x+y)^2}{xy} + 2 \frac{x+y}{\sqrt{xy}}$$
Now AM-GM implies that the first summand and second summand are both minimized at $x=y$.
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