Question

If x^2 + y^2 + z^2 = 3 and x, y, z are positive, prove that 2^1/x + 2^1/y + 2^1/z >= 6.

Answer 1

x=y=z=0x=y=z=0 then your answer is 33.

Suppose x=0,y=0x=0,y=0 then z=0z=0. So suppose x=0,y≠0,z≠0x=0,y≠0,z≠0. Then, y+z=0,y2=z,z2=yy+z=0,y2=z,z2=y. Subtracting the last to equations, (y−z)(y+z)=−(y−z)⟹y−z=0⟹y=z(y−z)(y+z)=−(y−z)⟹y−z=0⟹y=z. But as y+z=0y+z=0 we have y=z=0y=z=0.

So the answer is either 11 or 33.

Hope it helps you