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.

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.

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

Now it can be solved easily.

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Yup!That is one of the correct ways!

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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.

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@Harsh Shrivastava @Nihar Mahajan @Dev Sharma @Lakshya Sinha

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Any hint?

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Substitution.

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