Inspired by Toshit Jain

Solution 1:

First note that $y + z = 5 - x$ and $yz = 8 - x(y + z) = 8 - x(5 - x) = 8 - 5x + x^{2}$ . Then

$0 \le (y - z)^{2} = (y + z)^{2} - 4yz = (5 - x)^{2} - 4(8 - 5x + x^{2}) \Longrightarrow$

$0 \le 25 - 10x + x^{2} - 32 + 20x - 4x^{2} = -3x^{2} + 10x - 7 = -(3x^{2} - 10x + 7) \Longrightarrow$

$0 \le -(3x - 7)(x - 1) \Longrightarrow 1 \le x \le \dfrac{7}{3}$ .

So $M = \dfrac{7}{3}$ and $m = 1$ , and thus $\dfrac{a}{b} = \dfrac{7}{3} \Longrightarrow a + b = 7 + 3 = \boxed{10}$ .

The maximum for $x$ is achieved when $(x,y,z) = \left(\dfrac{7}{3}, \dfrac{4}{3}, \dfrac{4}{3}\right)$ and the minimum for $x$ is achieved when $(x,y,z) = (1,2,2)$ .

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Solution 2:

Alternatively, note that $x^{2} + y^{2} + z^{2} = (x + y + z)^{2} - 2(xy + yz + xz) = 5^{2} - 2 \times 8 = 9$ ,

which describes a sphere of radius $3$ .So the solution set to the given set of equations will be the intersection of this sphere and the plane $x + y + z = 5$ , which describes a circle. The circle is symmetric about the plane $y = z$ , (as $y$ and $z$ are interchangeable in the original system of equations), so both the maximum and minimum of $x$ will occur when $y = z$ . This transforms the given system of equations to

$x + 2y = 5 \Longrightarrow 2y = 5 - x$ and

$2xy + y^{2} = y(2x + y) = 8 \Longrightarrow 2y(4x + 2y) = 32 \Longrightarrow (5 - x)(4x + (5 - x)) = 32 \Longrightarrow$

$(5 - x)(3x + 5) = 32 \Longrightarrow -3x^{2} + 10x + 25 = 32 \Longrightarrow 3x^{2} - 10x + 7 = (3x - 7)(x - 1) = 0$ ,

from which we see that $M = \dfrac{7}{3}$ and $m = 1$ as found before.