Show inequality $a\sqrt{b - 1} + b\sqrt{a - 1} \le ab$ Given numbers $a$ and $b$; $a, b \ge 1.$
I'm trying to prove 
$$a\sqrt{b-1} +b\sqrt{a - 1} \le ab.$$
Also conditions for turning it to equality.
I tried to use AM-GM to the $(a - 1)(b - 1)$, which less than ab, but got nothing.
Applying AM-GM to $a\sqrt{b-1}$ and $b\sqrt{a - 1}$ also didn't give me some result. How can i do it?

Set $x=a-1,y=b-1.$
$$\begin{aligned}LHS&=(x+1)\sqrt y +(y+1)\sqrt x\\\\ &=\frac{(x+1)(y+1)}{2}\cdot \left[\frac{2}{\sqrt y + \frac{1}{\sqrt y}}+\frac{2}{\sqrt x + \frac{1}{\sqrt x}}\right]\\\\&\leq(x+1)(y+1)=RHS,\end{aligned}$$
because $t+{1\over t} \geq 2\;$ for all $t>0.$
Equality occurs when $t=1$ or, in terms of $a$ and $b,$ when $a=b=2.$ 
