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?
 A: We have $\sqrt{x-1}\leq x/2$: this is because the LHS is concave and tangent-line equality occurs at $x=2$. So we have
$$
a\sqrt{b-1}+b\sqrt{a-1}\leq a(b/2)+b(a/2) =ab
$$
A: 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 $a=b=2.$ 
A: Even though this problem have elementary solutions, I am a fan of using trigonometric methods to inequalities. Given $a, b\ge1,$ we can find unique $\phi,\psi\in\left[0, \frac{\pi}2\right)$ such that $a=\sec^2\phi$ and $b=\sec^2\psi.$ Since $$1+\tan^2\theta=\sec^2\theta$$ for any angle (other than odd multiples of $\pi/2$), LHS of the required inequality reduces to $$\sec^2\phi\tan\psi\,+\sec^2\psi\tan\phi\,=\dfrac12\sec^2\phi\sec^2\psi(\sin 2\psi\,+\sin 2\phi).$$ Now using the fact that $0\le\sin\theta\lt1$ for any $\theta\in\left[0, \pi\right),$ we get the solution.
