Show that $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$ How can one show that $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$
Assuming  that :
$\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$ 
So
$(\sqrt{x-1}+\sqrt{y-1})^2\leq xy$
$\sqrt{(x-1)(y-1)} \leq xy-x-y+2$
$ (y-1)(x-1)+3 \leq \sqrt{(x-1)(y-1)}$
Here I'm stuck !

Starting with “Assuming that $X$” when $X$ is the thing to be proved is not the right thing to do. You can make your computations easier if you set
$$
t=\sqrt{x-1},\quad u=\sqrt{y-1}
$$
so $x=t^2+1$ and $y=u^2+1$. The inequality to be proved becomes
$$
t+u\le\sqrt{(t^2+1)(u^2+1)}
$$
that has a single radical. Since $t+u\ge0$, the inequality is equivalent to
$$
(t+u)^2\le(t^2+1)(u^2+1)
$$
that is,
$$
t^2+2tu+u^2\le t^2u^2+t^2+u^2+1
$$
and so equivalent to
$$
0\le t^2u^2-2tu+1
$$
(by transporting terms to the right-hand side) which becomes
$$
0\le(tu-1)^2
$$
which is true.
Don't forget the $2$ when you square! You basically did $(a+b)^2=a^2+ab+b^2$ and this is where you got stuck.
