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Algebra Level 4

If x x , y y and z z are positive reals, such that x y z = 2 xyz = 2 then find the minimum of x 4 + 4 y 2 + 4 z 4 x^4 + 4y^2 + 4z^4

This problem is original


The answer is 16.

Dev Sharma by Dev Sharma , 20, India

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4 solutions

Deepak Kumar
Nov 5, 2015

Simple application of AM-GM.Just split 4y^2 as 2y^2+2y^2 and we get 16 as the minimum required value.

6 Helpful 0 Interesting 0 Brilliant 0 Confused

What are the values of x,y and z?

riccardo lombardi - 5 years, 7 months ago
Atul Shivam
Nov 12, 2015

simple question

apply A M − G M AM-GM inequality x 4 + 2 y 2 + 2 y 2 + 4 z 4 4 ≥ x 4 × 2 y 2 × 2 y 2 × 4 z 4 4 \frac {x^4+2y^2+2y^2+4z^4}{4} \geq \sqrt[4]{x^4×2y^2×2y^2×4z^4}

x 4 + 2 y 2 + 2 y 2 + 4 z 4 ≥ 4 x 4 × 2 y 2 × 2 y 2 × 4 z 4 4 x^4+2y^2+2y^2+4z^4 \geq 4\sqrt[4]{x^4×2y^2×2y^2×4z^4}

x 4 + 2 y 2 + 2 y 2 + 4 z 4 ≥ 4 × 2 x y z x^4+2y^2+2y^2+4z^4\geq 4×2xyz x 4 + 2 y 2 + 2 y 2 + 4 z 4 ≥ 16 x^4+2y^2+2y^2+4z^4\geq 16 so the correct answer is 16 \boxed{16}

3 Helpful 0 Interesting 1 Brilliant 0 Confused

Did the exact same

Aditya Kumar - 5 years ago

x 4 + 4 y 2 + 4 z 4 = x 4 + 2 y 2 + 2 y 2 + 4 z 4 By AM-GM inequality, x 4 + 2 y 2 + 2 y 2 + 4 z 4 ≥ 4 ⋅ x 4 ⋅ 2 y 2 ⋅ 2 y 2 ⋅ 4 z 4 4 = 16 \begin{aligned} x^4+4y^2+4z^4&=x^4+2y^2+2y^2+4z^4 \\ \text{By AM-GM inequality,} x^4+2y^2+2y^2+4z^4 &\geq 4 \cdot \sqrt[4]{x^4\cdot 2y^2\cdot 2y^2\cdot 4z^4}=\boxed{16}\\ \end{aligned}

0 Helpful 0 Interesting 0 Brilliant 1 Confused
Tamir Dror
Nov 15, 2015

It's obvious x=y=squared 2 z=1 So 16

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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