JBMO2020 18.04.2020 11:35 Prove that if $x, y, z$ are reals, then $x^2(3y^2+3z^2-2yz)=>yz(2xy+2xz-yz)$
RagvaloD 18.04.2020 15:06 $3x^2y^2+3x^2z^2+y^2z^2 \geq 2yz (xy+xz)+2x^2yz$ $3x^2y^2+3x^2z^2+y^2z^2 \geq 2(x^2y^2+2x^2z^2)+y^2z^2+2x^2yz \geq (xy+xz)^2+y^2z^2+2x^2yz \geq 2yz(xy+xz)+2x^2yz$ Equality if $xy=xz, yz=xy+xz \to y=z=2x$