assume that (x-y)^2*3/4 is max 4x^2+4y^2+4z^2 >= 4xy+4yz+4xz+3x^2 - 6xy+3y^2 x^2+2xy+y^2 + 4z^2 >= 4yz+4xz (x+y-2z)^2>= 0

Quote: assume that $ \frac{(x-y)^{2}\times 3}{4}$ is max $ 4x^{2}+4y^{2}+4z^{2} \geq 4xy+4yz+4xz+3x^{2} - 6xy+3y^{2}$ $ x^{2}+2xy+y^{2} + z^{2} \geq 4yz+4xz$ $ (x+y-2z)^{2}\geq 0$ I jut made small typo to this very nice and simple solution Thanks vg23nov.

assume that $ \frac{3}{4}(x-y)^2$ is the maximum rewrite the LHS as : $ \frac{1}{2}(x-y)^2+\frac{1}{2}(y-z)^2+\frac{1}{2}(x-z)^2$ so we need to prove that $ (y-z)^2+(x-z)^2 \geq \frac{1}{2}(x-y)^2$ it is equivalent to $ a^2+b^2 \geq \frac{1}{2}(a+b)^2$ which is true , if we let $ a=y-z ; b=z-x$

really an easy one)

delegat wrote: For arbitrary reals $ x$, $ y$ and $ z$ prove the following inequality: $ x^{2} + y^{2} + z^{2} - xy - yz - zx \geq \max \{\frac {3(x - y)^{2}}{4} , \frac {3(y - z)^{2}}{4} , \frac {3(y - z)^{2}}{4} \}$ very inteligent problem