1
$\begingroup$

$x+\frac{1}{y}=10$; $y+\frac{1}{z}=10$; $z+\frac{1}{x}=10$;

What is the highest possible value of z?

  • 0
    Are you missing a $10$ in front of $x$ at the beginning?And what have you tried – 2017-01-21
  • 0
    No, I am not missing it, I know these kind of inequalities need a special way of solution, but I don't quite remember it – 2017-01-21
  • 0
    are they positive reals or what are they? – 2017-01-21
  • 0
    prove that follows from the System $$x=y=z$$ – 2017-01-21
  • 0
    Why does this question mention inequalities? This looks like system of $3$ equations in $3$ variables. – 2017-01-21

2 Answers 2

2

$x=\frac{1}{10-z}$ and $1=y\left(10-\frac{1}{10-z}\right)$,

which gives $y=\frac{10-z}{99-10z}$.

Hence, $$\frac{(10-z)z}{99-10z}+1=10z$$ or $$z^2-10z+1=0,$$ which gives the answer: $5+\sqrt{24}$.

0

As you put it, it's an equation with two solutions. The one with the highest z has $z=5+2\sqrt 6$