1,cho x,y,z,t thoa man $\left\{\begin{matrix} x+y+z+t=0 & & \\ x^2+y^2+z^2+t^2=3 & & \end{matrix}\right.$
tim max xyzt
$GT\Rightarrow \left\{\begin{matrix} y+z+t=-x \\ y^2+z^2+t^2=3-x^2 \end{matrix}\right.$
Áp dụng BĐT $Cauchy-Schwarz$, ta được:
$y^2+z^2+t^2\geq \frac{(y+z+t)^2}{3}\Rightarrow 3-x^2\geq \frac{x^2}{3}\Rightarrow 9\geq 4x^2\Rightarrow x\leq \frac{3}{2}$
$Cmtt$, ta có:
$x,y,z,t\leq \frac{3}{2}\Rightarrow xyzt\leq \frac{81}{16}$
$\color{red}{\mathrm{\text{How I wish I could recollect, of circle roud}}}$
$\color{red}{\mathrm{\text{The exact relation Archimede unwound ! }}}$