IMO SHORTLIST, 1987

[opincariumihai]

Daca \( x,y,z \) sunt numere reale cu \( x^2+y^2+z^2=2 \), demonstrati ca \( x+y+z \leq 2+xyz \).

[Mateescu Constantin]

\( \left\|\ \begin{array}{cc}
x\ ,\ y\ ,\ z\ \in\ \mathbb{R} \\\\\\\\
x^2+y^2+z^2=2\ (*)\ \end{array}\right\|\ \Longrightarrow\ x+y+z\le 2+xyz\ \ \ \ \) (IMO shortlist 1987) .

Avem \( \ \ 4-(xyz-x-y-z)^2=4-(x+y+z)^2+2xyz(x+y+z)-x^2y^2z^2= \)

\( =^{(*)}\ \frac 14\left\[2(x^2+y^2+z^2)^3-(x^2+y^2+z^2)^2(x+y+z)^2+4xyz(x+y+z)(x^2+y^2+z^2)-4x^2y^2z^2\right\] \)

\( =\ \frac 14\left\[(x^2+y^2+z^2)^3-2(xy+yz+zx)(x^2+y^2+z^2)^2+4xyz(x+y+z)(x^2+y^2+z^2)-4x^2y^2z^2\right\] \)

\( =\ \frac 14\left\[(x^2+y^2+z^2-2yz)(x^2+y^2+z^2-2zx)(x^2+y^2+z^2-2xy)+4x^2y^2z^2\right\] \)

\( =\ \frac 14\left\{\left\[x^2+(y-z)^2\right\]\left\[y^2+(z-x)^2\right\]\left\[z^2+(x-y)^2\right\]+4x^2y^2z^2\right\}\ \ge\ 0 \)

Prin urmare \( 4-(xyz-x-y-z)^2\ge 0\ \Longleftrightarrow\ |xyz-x-y-z|\le 2\ \Longleftrightarrow\ \overline{\underline{\left\|\ x+y+z\ \le\ 2+xyz\ \right\|}}
\ \le\ 4+x+y+z \)