Indian Statistical Institute B.Math & B.Stat : Inequality

If $a,b,c \in (0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)} \geq 8.$ Let $p = 1-a, q = 1-b, r = 1-c$. $p+q+r= 3-(a+b+c)=1$.Clearly $p,q,r$ are positive. Substituting in the given inequality, it transforms to $\frac{(1-p)(1-q)(1-r)}{pqr} \geq 8 \iff {(1-p)(1-q)(1-r)} \geq 8{pqr}$ $\iff 1-(p+q+r)+qr+rp+pq-pqr \geq 8{pqr} \iff qr+rp+pq-pqr \geq 8{pqr}$ $\iff qr+rp+pq \geq 9{pqr} \iff \frac{1}{p}+\frac{1}{q}+\frac{1}{r} \geq 9$ Thus proving the above inequality reduces to proving $\frac{1}{p}+\frac{1}{q}+\frac{1}{r} \geq 9$ subjected to $p+q+r=1$ Since $p,q,r$ are positive, apllying $A.M \geq H.M$ we have $\frac{p+q+r}{3} \geq \frac{3}{\frac{1}{p}+\frac{1}{q}+\frac{1}{r}}$ Noting $p+q+r=1$, we have $\frac{1}{p}+\frac{1}{q}+\frac{1}{r} \geq 9$