Inspired by Harsh Shrivastava

$x+y+z=4 \\ x^{2}+y^{2}+z^{2}=6 \\ xy+yz+zx=5$

Let $x,y,z$ be the roots of some polynomial $P(a)$ . Applying discrimination of cubic equation we will get an equation as.

$0 \ge 27d^{2}-104d+100$

Where $d=xyz$ . We see the above value attains it maximum at $0$ . Therefore the roots of equation will give the maximum value $xyz$ . The roots of above equation are $2, \frac{100}{54}$ . But only 2 is acceptable. How? That depends upon you.

My approach is similar to Lakshya's, but I avoid the somewhat tedious work with the discriminant.

By Viete, we have the polynomial $f(t) = t^3-4t^2+5t-p$ whose roots are $x,y,z$ , where $p=xyz$ . This polynomial has the local maximum $f(1)=2-p$ . If $p>2$ , we have only one real solution, but for $p=2$ we get the solutions $1,1,2$ for $x,y,z$ , in any order. Thus the maximum product is $p=\boxed{2}$ .

When I get to xy+xz+yz=5, I replace y+z by 4-x, and yz by d/x giving the equation (after rearranging): d=x^3-4x^2+5x Differentiating to get max d gives x=1 or 5/3 giving d = 2 or 1.85... Hence max d is 2.

Umm....I used quite an unorthodox way.I assumed that x^2,y^2,z^2 would be positive(since they aren't complex).And the only positive numbers whose squares add up to 6 were 2,1,1 . On top of that 2+1+1 was equal to 4.

Ha ha.., i thought it in a simpler way...... I thouht what values of xyz we can put to satisfy eq. Putting two of them as 1 and one as 2....

Hit and trail rocks

Apply AM>=GM on both the equations.Then select the intersection of the inequalities and then GIF.

x+ y = 4 - z

$x^2 + y^2 = 6 - z^2$

using cauchy schwarz, then its easy

The question doesn't make sense.x=y=z yields 2.37 as the value of the product.