Question 461593
Consider the vectors (x, y, z) and ( {{{1/sqrt(3)}}}, {{{1/sqrt(3)}}},{{{1/sqrt(3)}}}).  
Then direct application of the Cauchy-Schwartz Inequality gives:

{{{(x^2 + y^2 + z^2)*((1/sqrt(3))^2 +(1/sqrt(3))^2 + (1/sqrt(3))^2) >= (x/sqrt(3) + y/sqrt(3) + z/sqrt(3))^2}}}

{{{(x^2 + y^2 + z^2)*1 >= (1/3)(x+y+z)^2}}}

{{{3x^2 + 3y^2 + 3z^2 >= x^2 + y^2 + z^2  + 2xy + 2xz + 2yz}}}

{{{2x^2 +2y^2 + 2z^2 >=  2xy + 2xz + 2yz}}}
{{{x^2 +y^2 + z^2 >=  xy + xz + yz}}}.

we don't even need the non-negativity conditions for x, y, and z.