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 \(\frac{1}{p}+\frac{1}{q}+\frac{1}{r} \geq 9 \)