Suppose \(a,b,c\) be 3 positive reals such that

\(a+b+c≥\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)

Show that \(\frac{a^{3}c}{b(c+a)}+\frac{b^{3}a}{c(a+b)}+\frac{c^{3}b}{a(b+c)}≥\frac{3}{2}\).

I cannot still solve this problem.Please,someone help me .

Suppose \(a,b,c\) be 3 positive reals such that

\(a+b+c≥\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)

Show that \(\frac{a^{3}c}{b(c+a)}+\frac{b^{3}a}{c(a+b)}+\frac{c^{3}b}{a(b+c)}≥\frac{3}{2}\).

I cannot still solve this problem.Please,someone help me .

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TopNewest\[\displaystyle 2(a+b+c)\left( \frac{a}{b} +\frac{b}{c} +\frac{c}{a} \right) \left( \sum_{\text{cyc}}\frac{a^3c}{b(c+a)} \right)\]

\[\displaystyle \stackrel{\text{Hölder}}\ge (a+b+c)^3\stackrel{\text{?}}\ge 3(a+b+c)\left( \frac{a}{b} +\frac{b}{c} +\frac{c}{a} \right) \]

\[\displaystyle 3\left( \frac{a}{b} +\frac{b}{c} +\frac{c}{a} \right)\le 3(a+b+c)\stackrel{\text{?}}\le (a+b+c)^2\]

\[\displaystyle a+b+c\ge \frac{a}{b} +\frac{b}{c} +\frac{c}{a}\stackrel{\text{AM-GM}}\ge 3.\:\:\square\] – Mathh Mathh · 3 years, 1 month ago

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i think you should try to use AM-GM inequality – Akash Deep · 3 years, 1 month ago

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– Souryajit Roy · 3 years, 1 month ago

I have tried...but it was of no useLog in to reply

search up holder's inequality – Blah Blah · 3 years, 1 month ago

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