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.

#Algebra #Inequalities #NMTC

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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 - 6 years, 7 months ago

Amazing answer! Thank you!

Siddharth G - 6 years, 7 months ago

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 - 6 years, 5 months ago

@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 - 6 years, 5 months ago

@Siddharth G – You're right. Just skipped over that for some reason ^o^

Josh Banister - 6 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$