How to prove $\sqrt{\frac{bc}{a+bc}}+\sqrt{\frac{ca}{b+ca}}+\sqrt{\frac{ab}{c+ab}}\geq 1$ Let $a,b,c>0,a+b+c=1$, show that
$$\sqrt{\dfrac{bc}{a+bc}}+\sqrt{\dfrac{ca}{b+ca}}+\sqrt{\dfrac{ab}{c+ab}} \geq 1$$
I tried
$$\dfrac{bc}{a+bc}=\dfrac{bc}{a(a+b+c)+bc}=\dfrac{bc}{(a+b)(a+c)}$$
It suffice to show that
$$\sqrt{\dfrac{bc}{(a+b)(a+c)}}+\sqrt{\dfrac{ca}{(b+a)(b+c)}}+\sqrt{\dfrac{ab}{(c+a)(c+b)}}\geq 1$$
 A: Let $a=\frac{1}{x}$, $b=\frac{1}{y}$ and $c=\frac{1}{z}$.
Hence, we need to prove that$\sum\limits_{cyc}\frac{x}{\sqrt{(x+y)(x+z)}}\geq1$.
By AM-GM and C-S $\sum\limits_{cyc}\frac{x}{\sqrt{(x+y)(x+z)}}\geq\sum\limits_{cyc}\frac{2x}{2x+y+z}\geq\frac{2(x+y+z)^2}{\sum\limits_{cyc}(2x^2+2xy)}\geq1$
A: As $0<a,b,c<1$, we can find $0<A,B,C<\pi/2$ such that $a=\sin^2A, b=\sin^2B$ and $c=\sin^2C$.
Then, we have $\sin^2A + \sin^2B + \sin^2C=1$.
If $A+B+C<\pi/2$, $\sin(A+B) < \sin(\pi/2 - C) = \cos(C)$. Squaring both sides, $\cos^2C > \sin^2A\cos^2B + \sin^2B\cos^2A + 2\sin A\sin B\cos A\cos B$ $= \sin^2A + \sin^2B -2\sin^2A\sin^2B + 2\sin A\sin B\cos A\cos B$
$= \sin^2A + \sin^2B + 2\sin A \sin B\cos (A+B)$
$\geq \sin^2A + \sin^2B$ as $A+B < A+B+C < \pi/2$ and so $\cos (A+B)>0$
Thus, we have $\sin^2A+\sin^2B+\sin^2C < 1$ contradicting the given condition. Hence, we must have $A+B+C \geq \pi/2$.
Proceeding with the inequality,
$$\sqrt{\frac{bc}{(1-b)(1-c)}}+\sqrt{\frac{ab}{(1-a)(1-b)}}+\sqrt{\frac{ca}{(1-c)(1-a)}}$$
$$=\tan A\tan B + \tan B\tan C + \tan C\tan A$$
$$=\frac{\sin A\sin B\cos C + \sin B\sin C\cos A + \sin C\sin A\cos B}{\cos A\cos B\cos C}$$
$$=\frac{\cos A\cos B\cos C-\cos(A+B+C)}{\cos A\cos B\cos C} = 1 - \frac{\cos(A+B+C)}{\cos A\cos B\cos C} \geq 1$$
because $A+B+C\geq \pi/2$ implies $\cos(A+B+C)\leq 0$. Thus the inequality stands proven.
