Minimum of sum of squares

Algebra Level pending

Let x x , y y and z z be positive real numbers such that x + 2 y + z = 1 x+2y+z=1 . What is the minimum value of x 2 + y 2 + z 2 { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } ?

Give your answer to two decimal places.


The answer is 0.17.

Anthony Muleta by Anthony Muleta , 23, Australia

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1 solution

Anthony Muleta
Feb 10, 2015

By the Cauchy-Schwarz inequality, we have

( x 2 + y 2 + z 2 ) ⋅ ( 1 2 + 2 2 + 1 2 ) ≥ ( x + 2 y + z ) 2 ⇒ 6 ( x 2 + y 2 + z 2 ) ≥ 1 ⇒ x 2 + y 2 + z 2 ≥ 1 6 \left( { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right) \cdot \left( { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 1 }^{ 2 } \right) \ge { \left( x+2y+z \right) }^{ 2 }\\ \Rightarrow 6\left( { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right) \ge 1\\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }\ge \frac { 1 }{ 6 }

Hence, the minimum value is 0.17 \boxed { 0.17} .

0 Helpful 0 Interesting 1 Brilliant 0 Confused

I used Lagrange Multipliers for this question. I used this method because I need to study it for my Calc 2 exam. I liked your solution cause it is simple!

Victor Paes Plinio - 1 year, 6 months ago

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