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

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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

`(x+y)(y+z)(z+x)gt=(2sqrt(xy))*(2sqrt(yz))*(2sqrt(zx))`

`(x+y)(y+z)(z+x)gt=8(sqrt(x^2*y^2*z^2))`

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 lastĀ line proves that the inequality is checked.

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