If x, y and z are real numbers such that x + y + z = 5 and xy + yz + zx = 3, what is the largest value that x can have?

The given equations are  x + y + z = 5 -- (1) , xy + yz + zx = 3 -- (2)
xy + yz + zx = 3
x(y + z) + yz = 3
=> x ( 5 -x ) +y ( 5 - x - y) = 3
=> $$ - y^2 - y (5 -x) - x ^2 + 5x = 3$$
=> $$ y^2 + y (x-5) + ( x ^2 - 5x +3)  = 0 $$
The above equation should have real roots for y, => Determinant >= 0

=>$$b^2-4ac>0$$

=> $$ ( x - 5)^2 - 4(x ^2 - 5x +3 ) \geq 0$$
=>  $$ 3x^2 -10x - 13 \leq 0$$
=>  $$ -1 \leq x \leq \frac{13}{3}$$
Hence maximum value x can take is $$\frac{13}{3}$$, and the corresponding values for y,z are $$\frac{1}{3},\frac{1}{3}$$