$\sqrt{1+y^{2}}+ \sqrt{2y}\leq \sqrt{2}(1+y)$
$\sqrt{1+z^{2}}+ \sqrt{2z}\leq \sqrt{2}(1+z)$
$\Rightarrow A\leq \sqrt{2}(3+x+y+z)+ (3-\sqrt{2})(\sqrt{x}+\sqrt{y}+\sqrt{z})\leq \sqrt{2}.6+(3-\sqrt{2})\sqrt{3(x+y+z)}= 9 + 3\sqrt{2}$
dấu = xảy ra khi : x=y=z=1
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