Question

x,y,z>0.then show that (x+y+z)(1/x+1/y+1z(x + y + z)>9

Answer 1

Step-by-step explanation:

Simple. You can solve it with the identities taught in grade IX.

1/x + 1/y +1/z = 0

yz + xz + xy/ xyz = 0

yz + xz + xy = 0 …(I)

As given, x + y + z = 9 …(II)

Squaring both sides,

(x + y + z)² = (9)²

x² + y² + z² + 2xy + 2yz + 2xz = 81

x² + y² + z² + 2(xy + yz + xz) = 81

x² + y² + z² + 2(0) = 81 [By (I)]

x² + y² + z² = 81 … (III)

Now, we know that x³ + y³ + z³ - 3xyz = (x+y+z){x² + y² + z² - (xy + yz + zx)}

Substituting (I), (II) and (III) in the above equation:

x³ + y³ + z³ - 3xyz = (9)(81 - 0)

x³ + y³ + z³ - 3xyz = 729

And that's your answer.

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