Minimum Value

Algebra Level 5

Let x , y , z x,y,z be positive real numbers such that x 2 + y 2 + z 2 + 2 x y z = 4 x^2 + y^2 + z^2 +2xyz = 4 .

Find the minimum value of x 2 + y 2 + z 2 x^2 + y^2 + z^2 .


The answer is 2.4887.

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Fahim Shahriar Shakkhor by Fahim Shahriar Shakkhor , 24, Bangladesh

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1 solution

By the AM-GM inequality we know that x 2 + y 2 + z 2 ≥ 3 ∗ ( x y z ) 2 3 . x^{2} + y^{2} + z^{2} \ge 3*(xyz)^{\frac{2}{3}}. Thus the minimum value of x 2 + y 2 + z 2 x^{2} + y^{2} + z^{2} will occur when

3 ∗ ( x y z ) 2 3 + 2 x y z = 4 ⟹ 8 ( x y z ) 3 − 21 ( x y z ) 2 + 96 ( x y z ) − 64 = 0 , 3*(xyz)^{\frac{2}{3}} + 2xyz = 4 \Longrightarrow 8(xyz)^{3} - 21(xyz)^{2} + 96(xyz) - 64 = 0,

which has one real solution of x y z = 0.75561 xyz = 0.75561 to 5 5 decimal places. Thus the minimum value of x 2 + y 2 + z 2 x^{2} + y^{2} + z^{2} will be 4 − 2 ∗ 0.75561 = 2.48878 4 - 2*0.75561 = \boxed{2.48878} to 5 5 decimal places.

4 Helpful 0 Interesting 0 Brilliant 0 Confused

How did you get from 3 ( x y z ) 2 3 + 2 x y z = 4 → t o → 8 ( x y z ) 3 − 21 ( x y z ) 2 + 96 ( x y z ) − 64 = 0 ? 3(xyz)^{\frac{2}{3}} + 2xyz = 4 \rightarrow\ to \rightarrow \ 8(xyz)^3 -21(xyz)^2 + 96(xyz) -64 = 0\ ?

Curtis Clement - 6 years, 3 months ago

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Let a = x y z . a = xyz. Then we have 3 a 2 3 = 4 − 2 a . 3a^{\frac{2}{3}} = 4 - 2a. Now cube both sides to get

27 a 2 = 8 ( 2 − a ) 3 = 8 ( 8 − 12 a + 6 a 2 − a 3 ) = 64 − 96 a + 48 a 2 − 8 a 3 27a^{2} = 8(2 - a)^{3} = 8(8 - 12a + 6a^{2} - a^{3}) = 64 - 96a + 48a^{2} - 8a^{3}

⟹ 8 a 3 − 21 a 2 + 96 a − 64 = 0. \Longrightarrow 8a^{3} - 21a^{2} + 96a - 64 = 0.

This is a messy cubic polynomial so I used WolfraAlpha to find the root, but an application of Newton's Method would do the job as well.

Brian Charlesworth - 6 years, 3 months ago

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