Math Problem

[jimerax]

Okay, I'll show you the proof of this problem, it was hard.

Since x>0, y>0, z>0, then

yz/x+zx/y+xy/z >= 3^(1/2) ... (A)

<=>

(yz/x+zx/y+xy/z)^2 >= (3^(1/2) )^2

<=>

y^2*z^2/x^2 + z^2*x^2/y^2 + x^2*y^2/z^2 + 2(x^2 + y^2 + z^2) >= 3

Here we apply x^2 + y^2 + z^2 = 1,
<=>

y^2*z^2/x^2 + z^2*x^2/y^2 + x^2*y^2/z^2 - (x^2 + y^2 + z^2)
>= 0 ... (B)

(A) is equivalent to (B), so we need to prove (B).

Here we apply AM-GM inequality [(X+Y)/2 >= sqrt(XY)] provided that x>0, y>0, z>0,

(y^2*z^2/x^2 + z^2*x^2/y^2) / 2 >= sqrt(y^2*z^2/x^2 * z^2*x^2/y^2) = z^2

<=>

(y^2*z^2/x^2 + z^2*x^2/y^2) / 2 - z^2 >= 0 ... (C)

In the same way,
(y^2*z^2/x^2 + x^2*y^2/z^2) / 2 - y^2 >= 0 ... (D)
(z^2*x^2/y^2 + x^2*y^2/z^2) / 2 - x^2 >= 0 ... (E)

From (C), (D), (E), inequality (B) is proved.

so

yz/x+zx/y+xy/z >= 3^(1/2) // Q.E.D.

[lorianne]

i hate math