Cauchy inequality is (a12+a2 2+...) (b12+b22+...) ≥ (a1b1+a2b2+...) 2 Note: Cauchy inequality satisfies all real numbers. Therefore, there are:

(x^2+2y^2+3z^2)(9+2+ 1/3)≥(3x+2y+z)^2

-√( 18/ 17 * 34/ 13)≤3x+2y+z ≤√( 18/ 17 * 34/ 13)

That is, the minimum value is -2√3, if and only if X =-(9 √ 3)/ 17, y = (-3 √17, and Z = (-3)/ 17.

Don't forget to test the conditions for the establishment of the equal sign when finding the maximum value with inequality. This step is divided in the formal exam. )