Bonjour,
[tex](x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\\\\=1+2(xy+xz+yz)\\\\= 1+2\dfrac{(x+y)^2-(x-y)^2+(x+z)^2-(x-z)^2+(y+z)^2-(y-z)^2}{4}\\\\\leq 1+2\dfrac{(|x|+|y|)^2+(|x|-|y|)^2+(|x|+|z|)^2+(|x|-|z|)^2+(|z|+|y|)^2+(|z|-|y|)^2}{4}\\\\=1+2(|x|^2+|y|^2+|z|^2)=1+2(x^2+y^2+z^2)=3[/tex]En prenant la racine carrée, il vient
[tex]|x+y+z|\leq \sqrt{3}[/tex]
Merci
