1 Equation 3 variables!

Algebra Level 4

( 1 − x ) 2 + ( x − y ) 2 + ( y − z ) 2 + z 2 = 1 4 \left ( 1-x \right )^{2}+\left ( x-y \right )^{2}+\left ( y-z \right )^{2}+z^{2}=\frac{1}{4} Find all real numbers x , y , z x, y, z such that this equation is satisfied. If there exists only one ordered triplet ( x , y , z ) (x,y,z) that satisfies the equation. Then submit your answer as x 3 + 6 y + z . \sqrt{\frac{x}{3}}+6y+\sqrt{z}.

Bonus : Prove that there exists only one ordered triplet of ( x , y , z ) (x,y,z) .


The answer is 4.

Aaron Jerry Ninan by Aaron Jerry Ninan , 20, India

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1 solution

Aaron Jerry Ninan
Jul 31, 2016

By applying Power-mean inequality ( 1 − x ) 2 + ( x − y ) 2 + ( y − z ) 2 + z 2 4 ≥ [ ( 1 − x ) + ( x − y ) + ( y − z ) + z 4 ] 2 \frac{(1-x)^{2}+(x-y)^{2}+(y-z)^{2}+z^{2}}{4}\geq [\frac{(1-x)+(x-y)+(y-z)+z}{4}]^{2} Therfore we get ( 1 − x ) 2 + ( x − y ) 2 + ( y − z ) 2 + z 2 ≥ 1 4 (1-x)^{2}+(x-y)^{2}+(y-z)^{2}+z^{2}\geq \frac{1}{4} Since this is the minimum value of the equation,we get ( 1 − x ) = ( x − y ) = ( y − z ) = z (1-x)=(x-y)=(y-z)=z Now we have 3 equation and 3 variables,Solving for this we get, x = 3 4 x=\frac{3}{4} y = 1 2 y=\frac{1}{2} z = 1 4 z=\frac{1}{4} By substituting this in our required quantity we get the answer as 4

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