A Few Signs Switched

Relevant wiki: Muirhead Inequality

From the question, $x,y,z > 0$ . Hence, $2x > 0$ ; $x > -x$ . $x+y+z > -x+y+z$ .

$|x+y+z| = x+y+z > |-x+y+z|$ because if $x < y+z$ then $2x > 0$ ; $x > -x$ . $x+y+z > -x+y+z$ and if $x>y+z$ , then $x+y+z > x-y-z$ or $2(y+z) > 0$ .

Since $x+y+z > 0$ , we can divide both sides without changing the inequality sign:

$1 > \dfrac{|-x+y+z|}{x+y+z} > 0$ .

$1 > (\dfrac{-x+y+z}{x+y+z})^2 >0$ .

Similarly, we will also obtain $1 > (\dfrac{x-y+z}{x+y+z})^2>0$ and $1 > (\dfrac{x+y-z}{x+y+z})^2>0$ .

Adding altogether, we will get $3 > (\dfrac{-x+y+z}{x+y+z})^2 + (\dfrac{x-y+z}{x+y+z})^2 + (\dfrac{x+y-z}{x+y+z})^2$ .

As a result, we have the least upper bound, or the supremum of $S$ equals $a=3$ .

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Then according to Muirhead's Inequality , the sequence (2,0) majorizes (1,1), and so $x^2 + y^2 \geq 2xy$ . Similarly, $y^2 + z^2 \geq 2yz$ and $z^2 + x^2 \geq 2xz$ .

Adding altogether we will obtain: $2(x^2 + y^2 + z^2) \geq 2(xy + yz + zx)$ .

Then $3(x^2 + y^2 + z^2) \geq (x^2+y^2+z^2)+2(xy + yz + zx) = (x+y+z)^2$ .

Now since $(-x+y+z)^2 + (x-y+z)^2 + (x+y-z)^2 = 3(x^2+y^2+z^2) - 2(xy + yz +za)$ and $2(x^2 + y^2 + z^2) \geq 2(xy + yz + zx)$ , by adding up both sides, we will get: $(-x+y+z)^2 + (x-y+z)^2 + (x+y-z)^2 \geq x^2+y^2+z^2$

Therefore, $3[(-x+y+z)^2 + (x-y+z)^2 + (x+y-z)^2] \geq (x+y+z)^2$ .

Dividing by $3(x+y+z)^2$ both sides, we will get: $(\dfrac{-x+y+z}{x+y+z})^2 + (\dfrac{x-y+z}{x+y+z})^2 + (\dfrac{x+y-z}{x+y+z})^2 \geq \dfrac{1}{3}$ .

Thus, the infimum of $S$ is $b=\dfrac{1}{3}$ .

Finally, $\dfrac{a}{b} = 3\times 3 = \boxed{9}$ .