Maximum value for x

From the identity $(x+y+z)^2 = x^{2}+y^{2}+z^{2}+2(xy+yz+zx)$ , and substituting the values given , we get $x^{2}+y^{2}+z^{2}=19 \\ \implies y^{2}+z^{2}=19-x^2 \\ Also, \hspace{0.25cm} y+z=5-x$ From Cauchy-Schwarz Inequality , $y^{2}+z^{2} \geq \frac{1}{2}(y+z)^{2}$ Putting the values of $y^{2}+z^{2}$ and $y+z$ , we get: $19-x^2 \geq \frac{1}{2}(5-x)^2$ On simplifying we are left with $3x^2-10x-13 \leq 0$ which implies $(3x-13)(x+1) \leq 0$ yielding maximum value as $\dfrac{13}{3}$ $\approx$ $\boxed{4.33}$

Note that for showing that the value is achievable, we substitute $x=\dfrac{13}{3}$ from which we get that $y=z=\dfrac 13$ . Hence it is indeed achievable.

Here, x+y+z=5; xy+yz+zx=3; x^2+ y^2+z^2=(x+y+z)^2-2(xy+yz+zz)=25-6=19; x(y+z)+zx=3; Applying AM-GM x(y+z)+(z^2+y^2)/2≥3; 2x(5-x)+(19-x^2)≥6; -3x^2+10x+19-6≥0; 3x^2-10x-13≥0; (3x-13)(x+1)≥0; -1≤x≤13/3; Therefore maximum value for x is 13/3; which is achievable as putting this value is the set of equation will yield positive value for both Z and Y