If \(a,b,c\) are real numbers such that,\(0<a,b,c<1,a+b+c=2\).Then prove that,\[\prod_{cyc}\dfrac{a}{1-a} \geq 8.\].I have two solutions,one of which has been provided by Surya Prakash.

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TopNewestSubstitute a= 1-p ; b=1-q and c=1-r.

Now it can be solved easily. – Harsh Shrivastava · 1 year ago

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– Adarsh Kumar · 1 year ago

Yup!That is one of the correct ways!Log in to reply

– Svatejas Shivakumar · 1 year ago

After the substitution, multipling the terms and simplyfy using \(p+q+r=1\), we get \(pq+qr+rp \ge 9pqr\). Then divide the both sides of theb equation by \(pqr\). The inequality will hold by AM-HM inequality.Log in to reply

– Adarsh Kumar · 1 year ago

Yup!That is correct!Log in to reply

@Harsh Shrivastava @Nihar Mahajan @Dev Sharma @Lakshya Sinha – Adarsh Kumar · 1 year ago

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– Lakshya Sinha · 1 year ago

Any hint?Log in to reply

– Adarsh Kumar · 1 year ago

Substitution.Log in to reply