This chain of inequalities forms the foundation for many other classical inequalities. See how the four common "means" - arithmetic, geometric, harmonic, and quadratic - relate to each other.

\[\dfrac{1}{a^4}+\dfrac{1}{b^4}+\dfrac{1}{c^4}=1\]

Given the above equation for positive numbers \(a,b,c\).

Find the minimum value of

\[\dfrac{a^4b^4+a^4c^4+b^4c^4}{a^3b^2c^3}\]

If the minimum value of the above is \(x\), input your answer as \(\lfloor 100x \rfloor\).

This is part of the set Trevor's Ten

**Details and Assumptions**

The answer is not \(300\).

It is indeed \(a^3 b^2 c^3\) and not \(a^3 b^3 c^3 \)

\[\sum_{k=1}^{n} x_k^2 \le \sum_{k=1}^n \dfrac{1}{x_k^2}\]

Given that \(x_1,x_2, \ldots,x_n\) are positive reals whose sum is \(n\), find the largest integer \(n\) such that the inequality above always holds true.

If you think all positive integers \(n\) make the inequality hold true, enter 0 as your answer.

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