Question

If x + y + z = 0 and xyz = 1, then prove:
$\frac{1}{1 + {x}^{ 3} } + \frac{1}{1 + {y}^{3} } + \frac{1}{1 + {z}^{ 3} } = 1$
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Answer 1

Answer:

If x+y+z=1

then

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

and

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

and

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

Adding these we get

1/x+1/y+1/z=3+x/y+y/x+x/z+z/x+y/z+z/y

Let f(x)=x/y+y/x

This has turning points when 0=1/y−y/x2

ie when x=y

. This is a minimum as we can see that there is no maximum as x goes towards zero from above!

So f(x)=2

at its minimum!

Similarly for the other two pairs so that

1/x+1/y+1/z=3+x/y+y/x+x/z+z/x+y/z+z/y≥3+2+2+2=9

as required.

In fact exactly the same proof works for the following generalisation of the question:

If x1+x2…+xn=A

Then 1/x1+1/x2…+1/xn≥n2/A

Step-by-step explanation: