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NMTC Problem 3b

If x,y,z are each greater than 1, show that

\(\frac { { x }^{ 4 } }{ { (y-1) }^{ 2 } } +\frac { { y }^{ 4 } }{ { (z-1 })^{ 2 } } +\frac { { z }^{ 4 } }{ { (x-1) }^{ 2 } } \ge 48\)

This a part of my set NMTC 2nd Level (Junior) held in 2014.

Note by Siddharth G
2 years, 6 months ago

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Firstly, we have, \( (a-2)^2 \geq 0 \implies a^2 -4a + 4 \geq 0 \implies a^2 \geq 4(a-1) \implies \frac{a^2}{a-1} \geq 4 \implies \frac{a^4}{(a-1)^2} \geq 16 \)

Now, By AM-GM,

\( \frac{x^4}{(y-1)^2} + \frac{y^4}{(z-1)^2} + \frac{z^4}{(x-1)^2} \geq 3\sqrt[3]{\frac{x^4}{(y-1)^2}* \frac{y^4}{(z-1)^2}* \frac{z^4}{(x-1)^2} } = 3\sqrt[3]{\frac{x^4}{(x-1)^2}* \frac{y^4}{y-1)^2}* \frac{z^4}{(z-1)^2} } \geq 3\sqrt[3]{16*16*16} = 3 * 16 = 48. \) Siddhartha Srivastava · 2 years, 6 months ago

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@Siddhartha Srivastava Amazing answer! Thank you! Siddharth G · 2 years, 6 months ago

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@Siddharth G Is the AM-GM step necessary? You could just say \(\frac{x^4}{(x-1)^4} \geq 16\) and so on for y and z and add the 3 inequalities together right? Josh Banister · 2 years, 4 months ago

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@Josh Banister But in the question, the denominator and the numerator are of different variables, which makes the AM-GM necessary to bring the denominator and the numerator with the same variables together. Thus \(\frac { { x }^{ 4 } }{ { (x-1) }^{ 2 } } \ge 16\) can only be used after the AM_GM step. Siddharth G · 2 years, 4 months ago

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@Siddharth G You're right. Just skipped over that for some reason ^o^ Josh Banister · 2 years, 4 months ago

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