lower bound on $xy/(x+y)?$ I am looking forward to a lower bound on $x*y/(x+y),$ where both of $x\geq 1$ and $y\geq 1$ are real.
The upper bound on $x*y/(x+y)$ is obvious, for example $x$ or $y$. But a lower bound is not very obvious to me. I want a simple lower bound, which is not very complex.
Could anyone pass some wisdom to me? 
Thanks a lot in advance..
 A: Flip it over. After some simplification, we get $\dfrac{1}{y}+\dfrac{1}{x}$. It is not hard to find an upper bound for this.
A: $$\frac{xy}{x+y}=\frac{1}{\frac{1}{x}+\frac{1}{y}}$$
So to compute lower bound on this expression is same as computing upper bound on $\frac{1}{x}+\frac{1}{y}$
Since, $x\geq 1\implies \frac{1}{x}\leq1$ and similarly $\frac{1}{y}\leq 1$
Thus, $$\frac{1}{x}+\frac{1}{y}\leq 2 $$ $$\implies \frac{xy}{x+y}\geq \frac{1}{2}$$
A: Another way:
$$ \frac{xy}{x+y}=\frac{(x-1)(y-1)}{x+y}+1-\frac{1}{x+y}\geq 1-\frac{1}{x+y}\geq \frac{1}{2}$$
because $x+y \geq 2 \Rightarrow -\frac{1}{x+y}\geq -\frac{1}{2}$
A: The expression is symmetric in $x,y$, so set without loss of generality $y\ge x$. Then you can redefine $y$ as $y=Y\cdot x$, with $Y\ge 1$. The expression becomes a direct product
$$\frac{xy}{x+y} =\frac{x^2Y}{x+xY} = x\cdot\frac{Y}{1+Y},$$
which is linear in x, so clearly $x=1$ is required for a minimum. The minumum of the function $\frac{Y}{1+Y}$ which slowly converges against unity is taken at $y=Y=1$ as well, where it then evaluates to $\frac{1}{2}$.
A: Thank everyone above. Inspired by your collective intelligence, I came up with a lower bound that I would like to use:
x*y/(x+y)\geq 1/2*min(x,y) 
