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 ! 
 A: $$f(x,y) = \sqrt{x+y} \cdot \left( \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} \right) =  \sqrt{x+y} \frac{\sqrt{x} + \sqrt{y}}{\sqrt{xy}}$$
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 $x=y$.
A: 
How would you prove that $f$ has a global minimum without using all the differential calculus tools?

We can demonstrate the existence of a global minimum without too much effort without actually finding it.
Since $\displaystyle \frac{1}{\sqrt{x}} = \sqrt{1/x}$ and since the square root function is multiplicative, we can simplify the function to $f(x, y) = \sqrt{1 + y/x} + \sqrt{1 + x/y}$.  We can further rewrite $f$ as $g(z) = \sqrt{1 + z} + \sqrt{1 + 1/z}$, where $z = x/y$.  Notice that $g$ blows up for small $z$ and for large $z$.  Therefore, if $g$ has a global minimum, it will necessarily be for values of $z$ in a closed interval $[a, b] \subset \mathbb{R}^+$, which is a compact subset of $\mathbb{R}$.  Since $g$ is continuous, the extreme value theorem guarantees a minimum inside this interval.
A: Using A.M.$ \geqslant $  G. M. twice, we have
$$
\begin{aligned}
f(x, y) &=\sqrt{x+y} \cdot\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right) \\
& \geqslant \sqrt{2 \sqrt{x y}} \cdot 2 \sqrt{\frac{1}{\sqrt{x}} \cdot \frac{1}{\sqrt{y}}}  \\
&=2 \sqrt{2}
\end{aligned}
$$
Hence $f(x,y)$ attains its minimum $ \Leftrightarrow x=y$.
