The minimum value of x yz x 4 + y 4 + z 2 for positive real number x , y , z is

By A.M. ≥ G.M. x 4 + y 4 ≥ 2 x 2 y 2 and 2 x 2 y 2 + z 2 ≥ 8 x yz . ⇒ x yz x 4 + y 4 + z 2 ≥ 8

Mathematics - Limits Question with Solution | TestHub

MathematicsLimitsMiscellaneous/MixedMedium2 minPYQ_2010

MathematicsMediumsingle choice

The minimum value of $\frac{x ^{4} + y ^{4} + z ^{2}}{x yz}$ for positive real number $x, y, z$ is

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Answer:

Solution:

By A.M. $\geq$ G.M. $x^{4} + y^{4} \geq 2 x^{2} y^{2}$ and $2 x^{2} y^{2} + z^{2} \geq 8 x yz$ . $\Rightarrow \frac{x ^{4} + y ^{4} + z ^{2}}{x yz} \geq 8$

Stream:BITSATSubject:MathematicsTopic:LimitsSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2010

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