If x,y,z are real positive numbers show that (x+y)(y+z)(z+x) >= 8xyz

Expert Answers
sciencesolve eNotes educator| Certified Educator

You need to remember that `(sqrtx - sqrty)^2 gt=0 =gt x - 2sqrt(xy) + y gt= 0 =gt x + y gt 2sqrt(xy)` .

`(sqrty - sqrtz)^2 gt= 0 =gt y+ z gt= 2sqrt(yz)`

`(sqrtz - sqrtx)^2 gt= 0 =gt z+x gt= 2sqrt(zx)`

Multiplying the inequalities yields:



Since x,y,z are positive numbers => `(sqrt(x^2*y^2*z^2)) = xyz`

(x+y)(y+z)(z+x)`gt=` 8xyz

Hence, the last line proves that the inequality is checked.

mubinak27 | Student

Prove that x/y+y/z+z/x>=(x+y+z)/(xyz)^1/3, if x,y,z are natural numbers