Simple Inequality 2

Algebra Level 4

Given that $a,b,c$ are non-negative real numbers, then $ab^2+2bc^2+3ca^2\ge kabc$ for some positive real $k$ . What is the largest possible value of $k$ ? Round to the nearest thousandth.

The answer is 5.451.

1 solution

Jordi Bosch
Oct 13, 2014

By AM-GM we have:

$ab^{2} + 2bc^{2} + 3ca^{2} \ge 3 \sqrt[3]{6(abc)^{3}} = 3\sqrt[3]{6}abc$

Equality will hold when:

$ab^{2} = 2bc^{2} = 3ca^{2}$

$(a,b,c) = (\sqrt[3]{\frac{4}{3}}, \frac{2}{\sqrt[3]{\frac{4}{3}}}, 1)$ is a triple that achieves the equality case.

That means $k = 3\sqrt[3]{6} \approx 5.451$

you are right but if we take a=1,b=1,c=1 then we find that k=6 so i think there is a problem in solution. can you tell me? please..........

hansraj sharma - 6 years, 7 months ago

5.4 is less than 6, and were finding the greatest value of K that works foreverything.

Trevor Arashiro - 6 years, 7 months ago

Hey, please tell me how could we find the values of triplet?

Anandhu Raj - 6 years, 5 months ago

Simple Inequality 2

The answer is 5.451.

1 solution

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