cho 3 số thực dương x,y,z thỏa mãn xyz=1 tìm gtln của biểu thức[tex] P = \frac{1}{2x^{^{2}} + y^{^{2}} + 3} + \frac{1}{2y^{2} + z^{2} +3} + \frac{1}{2z^{2} + x^{2} + 3}[/tex]
Ta có: $(x-y)^2+(x-1)^2\ge 0$
$\Rightarrow 2x^2+y^2+3\ge 2(xy+x+1)$
$\Rightarrow \dfrac{1}{2x^2+y^2+3}\le \dfrac{1}{2(xy+x+1)}$
Suy ra $P\le \dfrac{1}{2(xy+x+1)}+\dfrac{1}{2(yz+y+1)}+\dfrac{1}{2(zx+z+1)}$
$\le \dfrac{1}{2}\left(\dfrac{1}{xy+x+xyz}+\dfrac{1}{yz+y+1}+\dfrac{1}{zx+z+1}\right)$
$\le \dfrac{1}{2}\left(\dfrac{1}{x(y+1+yz)}+\dfrac{x}{x(yz+y+1)}+\dfrac{1}{zx+z+1}\right)$
$\le \dfrac{1}{2}\left(\dfrac{x+1}{x(y+1+yz)}+\dfrac{1}{zx+z+1}\right)$
$\le \dfrac{1}{2}\left(\dfrac{x+1}{xy(1+xz+z)}+\dfrac{xy}{xy(zx+z+1)}\right)$
$\le \dfrac{1}{2}\dfrac{x+1+xy}{xy(1+xz+z)}=\dfrac{1}{2}\dfrac{x+1+xy}{xyz(xy+x+1)}=\dfrac{1}{2}$
Dấu "=" xảy ra $\Leftrightarrow x=y=z=1$
Có gì khúc mắc e hỏi lại nhé <3
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