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 !
 A: There are some errors in your calculation, e.g. a missing factor 2 in
$$
(\sqrt{x-1}+\sqrt{y-1})^2 = x - 1 + y - 1 + 2\sqrt{x-1}\sqrt{y-1}
$$
and in the last step the inequality sign is in the wrong direction and
the number $3$ is wrong.

For $x \ge 1$, $y \ge 1$ you can square the inequality (since both
sides are non-negative):
$$
\sqrt{\mathstrut x-1}+\sqrt{\mathstrut y-1}\leq \sqrt{\mathstrut xy} \\
\Longleftrightarrow (x-1) + (y-1) + 2 \sqrt{\mathstrut x-1}\sqrt{\mathstrut y-1} \le xy \\
\Longleftrightarrow 0 \le xy - x - y + 2 - 2 \sqrt{\mathstrut (x-1)(y-1)} \\
\Longleftrightarrow 0 \le (x-1)(y-1) -  2 \sqrt{\mathstrut (x-1)(y-1)} + 1
$$
With $t := \sqrt{(x-1)(y-1)}$ the right-hand side is
$$
 t^2 - 2 t + 1 = (t-1)^2 \ge 0 \, .
$$
so that the inequality is true.
It follows also that equality holds if and only if $t = 1$, 
i.e. if $(x-1)(y-1) = 1$.
A: 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.
A: WLOG $\sqrt x=\sec A,\sqrt y=\sec B$ where $0<A,B<\dfrac\pi2$
$\sqrt{x-1}+\sqrt{y-1}=\tan A+\tan B=\dfrac{\sin(A+B)}{\cos A\cos B}\le\sqrt{\sec A\sec B}=?$
