Áp dụng BĐT Bunhiacopxki:
\(1.\sqrt{1+x^2}+1.\sqrt{2x}\le\sqrt{2\left(x+1\right)^2}=\sqrt{2}\left(x+1\right)\)
Tương tự: \(\sqrt{1+y^2}+\sqrt{2y}\le\sqrt{2}\left(y+1\right)\); \(\sqrt{1+z^2}+\sqrt{2z}\le\sqrt{2}\left(z+1\right)\)
\(\left(3-\sqrt{2}\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\le\left(3-\sqrt{2}\right)\sqrt{3\left(x+y+z\right)}=9-3\sqrt{2}\)
Cộng vế với vế:
\(A\le\sqrt{2}\left(x+y+z+3\right)+9-3\sqrt{2}=9+3\sqrt{2}\)
\(A_{max}=9+3\sqrt{2}\) khi \(x=y=z=1\)