Find the minimum of $x^2 + y^2 + z^2$ ,where $x, y, z$ are
real numbers such that $x^3 + y^3 + z^3 − 3xyz =1$
UK 2008: Exercise 1.97 (BOMC)
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sourav das
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You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
Re: UK 2008: Exercise 1.97 (BOMC)
Let, $x^2 + y^2 + z^2 = A$, $xy + yz + zx = B$
$1 = (x^3 + y^3 + z^3 - 3xyz)^2 = (x+y+z)^2 (x^2 + y^2 + z^2 -xy -yz -zx)^2$
$=(A+2B)(A-B)^2 \leq \left ( \frac{(A+2B) + (A-B) + (A-B)}{3} \right)^3 = A^3$, by A.M-G.M with equality when $A+2B=A-B$ or $B=0$
SO, minimum of $x^2 + y^2 + z^2$ is $1$.
$1 = (x^3 + y^3 + z^3 - 3xyz)^2 = (x+y+z)^2 (x^2 + y^2 + z^2 -xy -yz -zx)^2$
$=(A+2B)(A-B)^2 \leq \left ( \frac{(A+2B) + (A-B) + (A-B)}{3} \right)^3 = A^3$, by A.M-G.M with equality when $A+2B=A-B$ or $B=0$
SO, minimum of $x^2 + y^2 + z^2$ is $1$.
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