Sequence And Series Question 400
Question: The minimum value of $ \frac{x^{4}+y^{4}+z^{2}}{xyz} $ for positive real number x, y, z is
Options:
A) $ \sqrt{2} $
B) $ 2\sqrt{2} $
C) $ 4\sqrt{2} $
D) $ 8\sqrt{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] By $ A.M.\ge GM. $ $ x^{4}+y^{4}\ge 2x^{2}y^{2} $ and $ 2x^{2}y^{2}+z^{2}\ge \sqrt{8}xyz. $
$ \Rightarrow \frac{x^{4}+y^{4}+z^{2}}{xyz}\ge \sqrt{8} $
BETA
