Question

If x, y, z are distinct positive numbers, prove that (x + y) (y + z) (z + x) > 8 xyz.
Further if x + y + z = 1, show that (1 - x) (1 - y) (1 - 2) > 8xyz.​

Answer 1

Answer:

a. tan60 x tan30 = ----------

b. sin245 = -----------

c. sin30 = cos----

d. sin60 x cos30 = ----------

e. cos260 = ------

Answer 2

Answer:

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:

`(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 last line proves that the inequality is checked.

Step-by-step explanation: