[Đại 10] Bất đẳng thức

C

congchuaanhsang

Đặt $t=x+y+z \Longrightarrow xy+yz+xz=\dfrac{t^2-3}{2} $

Theo gt $0 \le xy+yz+xz \le 3 \Longrightarrow \sqrt{3}\le t \le 3$

$\Longrightarrow xy+yz+xz+\dfrac{4}{x+y+z}=\dfrac{t^2}{2}+\dfrac{4}{t}-\dfrac{3}{2}=\dfrac{(t-3)(3t^2+9t-8)}{6t}+\dfrac{13}{3}$

Dễ thấy $\dfrac{(t-3)(3t^2+9t-8)}{6t} \le 0$ với $t \in [\sqrt{3} ;3]$

$\Longrightarrow Max =\dfrac{13}{3} \Longleftrightarrow x=y=z=1$
 
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