RMO Inequality Practice

If $a,b,c \in \mathbb{R}^{+}$ , such that $abc=1$ , then prove that $\displaystyle\sum_{cyc} \dfrac{a}{b} \geq \sum_{cyc} a$

Instruction: Those who have complete solution ready , please don't post it immediately.Post it minimum after 3 days of posting this note or give a small hint but not the solution.

#Algebra #Inequality #ForRMO #PracticeProblem

Easy Math Editor

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Comments

Multiplying by $abc$ and making the right side homogenous, it suffices to prove: $a^2c + b^2a + c^2b \geq (abc)^{\frac{2}{3}}(a+b+c) = a^{\frac{5}{3}}b^{\frac{2}{3}}c^{\frac{2}{3}} + a^{\frac{2}{3}}b^{\frac{5}{3}}c^{\frac{2}{3}} + a^{\frac{2}{3}}b^{\frac{2}{3}}c^{\frac{5}{3}}$ By AM-GM you know: $a^2c + a^2c + ab^2 \geq 3a^{\frac{5}{3}}b^{\frac{2}{3}}c^{\frac{2}{3}}$ $b^2a + b^2a + bc^2 \geq 3a^{\frac{2}{3}}b^{\frac{5}{3}}c^{\frac{2}{3}}$ $c^2b + c^2b + ca^2 \geq 3a^{\frac{2}{3}}b^{\frac{2}{3}}c^{\frac{5}{3}}$ Add them and you are done.

Alan Yan - 5 years, 8 months ago

How and why did you develop such approach....It's not obvious at all.....Plz. provide some better suggestions...

Anubhav Mahapatra - 3 years, 8 months ago

i think we can make subsitutions... a = x/y, b = y/z and c = z/x

Dev Sharma - 5 years, 8 months ago

I recommend you to solve it completely and then post it.If you have some thoughts/ideas , you can post them after a complete solution is posted.