The minimum value of ( 8 x 2 + y 2 + z 2 ) ( x 1 + y 1 + z 1 ) 2 , ( x , y , z > 0 ) , is

2 + 1 + 1 2 ( 2 x ) 2 + y 2 + z 2 ≥ ( 2 + 1 + 1 2 ( 2 x ) + y + z ) 2 ≥ ( 2 x 2 + y 1 + z 1 2 + 1 + 1 ) 2 ⇒ ( 8 x 2 + y 2 + z 2 ) ( x 1 + y 1 + z 1 ) 2 ≥ 64 .

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MathematicsSequence & SeriesMeans / inequalityMedium2 minQB

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The minimum value of $(8 x^{2} + y^{2} + z^{2}) (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^{2}, (x, y, z > 0)$ , is

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Solution:

$\frac{2 ( 2 x ) ^{2} + y ^{2} + z ^{2}}{2 + 1 + 1} \geq (\frac{2 ( 2 x ) + y + z}{2 + 1 + 1})^{2} \geq (\frac{2 + 1 + 1}{\frac{2}{2 x} + \frac{1}{y} + \frac{1}{z}})^{2} \Rightarrow (8 x^{2} + y^{2} + z^{2}) (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^{2} \geq 64$ .

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Means / inequality

⏱ 2mℹ️ Source: QB

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