Question

Aratati ca$x^{2}$+$y^{2}$+$z^{2}$≥xy+xz+yz, (aici apare un A rasturnat) x,y,z ∈R

Answer 1

Inegalitatea este echivalenta cu:           (inmultesc cu 2)

$2 x^{2} +2 y^{2} + 2z^{2} \geq 2xy+2xz+2yz\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ 2 x^{2} +2 y^{2} +2 z^{2} -2xy-2xz-2yz \geq 0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ ( x^{2} -2xy+ y^{2}) +( y^{2}- 2yz+ z^{2})+( z^{2}-2xz+ x^{2}) \geq 0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ (x-y) ^{2} +(y-z) ^{2}+ (z-x)^{2} \geq 0$

Ceea ce este evident adevarat pentru ca $(x-y) ^{2} \geq 0~;~(y-z) ^{2} \geq 0~si~(z-x) ^{2} \geq 0.$