Question 398582
It is symmetric, but doesn't necessarily mean that x = y = z (we could prove that it's true if it is).


Without loss of generality, suppose {{{0 <= x <= y <= z}}} (we can make such an assumption because it's symmetric, but we must also consider negative solutions, if they exist). Then,


{{{2z^2 >= 2y^2 >= 2x^2}}}


{{{14 + xy >= 14 + xz >= 14 + yz}}} --> {{{xy >= xz >= yz}}} (using substitution)


Taking subsets of the inequality, we get {{{xy >= xz}}}, {{{xy >= yz}}}, {{{xz >= yz}}}, which implies {{{y >= z}}}, {{{x >= z}}}, {{{x >= y}}}. Since we already have {{{z >= y}}}, {{{z >= x}}}, {{{y >= x}}} from our first assumption, this implies {{{x = y = z}}}. Setting all variables equal to x, we get {{{2x^2 = 14 + x^2}}}, {{{x = sqrt(14)}}}, so ({{{sqrt(14)}}}, {{{sqrt(14)}}}, {{{sqrt(14)}}}) is a solution.


If we assume all x,y,z are negative, then by the same logic we get the ordered triple ({{{-sqrt(14)}}}, {{{-sqrt(14)}}}, {{{-sqrt(14)}}}) as another solution.


However, what happens if some of the numbers are positive and others are negative? In this case we can assume {{{0 < abs(x) <= abs(y) <= abs(z)}}}. Since the magnitudes of the squares of the numbers are still in order, we get {{{2z^2 >= 2y^2 >= 2x^2}}}, and the inequalities {{{y >= z}}}, {{{x >= z}}}, and {{{x >= y}}} similar to above. Now, suppose we assume that z is positive. If this is the case, x and y are both positive, and their magnitudes are larger, thus contradiction, so z is negative.


Suppose we subtract the first equation from the second to obtain


{{{2y^2 - 2x^2 = zx - yz = z(x-y)}}}. The left side is positive since the magnitude of y is larger than the magnitude of x. Since z is negative, it follows that x-y is also negative. This implies either:


x is positive, and since y has a magnitude larger than x, y is also positive. However this would not satisfy the inequality {{{x >= y}}} (unless x = y).


x is negative, and since y has a larger magnitude and x-y is less than zero, then y is positive. However this would not satisfy {{{x >= y}}} in any case (since none of the variables can be zero).


The only possibility is if {{{x = y}}} and {{{z < 0}}}. If we replace y with x we get the system:


{{{2x^2 = 14 + xz}}}
{{{2z^2 = 14 + x^2}}}


Subtracting the first equation from the second equation,


{{{2z^2 - 2x^2 = x^2 - xz = x(x - z)}}}


{{{2(z-x)(z+x) = x(x-z)}}} Cancel z-x


{{{2(z-x) = -x}}}


{{{2z - 2x = -x}}} --> {{{2z = x}}} --> {{{z = x/2}}}. Therefore we can replace y with x, z with x/2 in the first and third equations to obtain:


{{{2x^2 = 14 + x(x/2)}}} --> {{{(3/4)x^2 = 14}}}
{{{2(x/2)^2 = 14 + x^2}}} --> {{{(-1/2)x^2 = 14}}}


These two equations cannot be satisfied, since it implies {{{3/4 = -1/2}}} (note that x cannot be zero, it can be proven using contradiction that none of x,y,z can equal zero). Therefore there are no solutions in this case.


Thus, the only solutions are when x = y = z, or ({{{sqrt(14)}}}, {{{sqrt(14)}}}, {{{sqrt(14)}}}) and ({{{-sqrt(14)}}}, {{{-sqrt(14)}}}, {{{-sqrt(14)}}}).