Lời giải:
Áp dụng BĐT Bunhiacopxky:
\((\sqrt{1+x^2}+\sqrt{2x})^2\leq (1+x^2+2x)(1+1)\)
\(\Leftrightarrow \sqrt{1+x^2}+\sqrt{2x}\leq \sqrt{2}(x+1)\)
Hoàn toàn tt: \(\left\{\begin{matrix} \sqrt{1+y^2}+\sqrt{2y}\leq \sqrt{2}(y+1)\\ \sqrt{1+z^2}+\sqrt{2z}\leq \sqrt{2}(z+1)\end{matrix}\right.\)
Tiếp tục Bunhiacopxky:
\((\sqrt{x}+\sqrt{y}+\sqrt{z})^2\leq (x+y+z)(1+1+1)\)
\(\Rightarrow (2-\sqrt{2})(\sqrt{x}+\sqrt{y}+\sqrt{z})\leq (2-\sqrt{2})\sqrt{3(x+y+z)}\)
Cộng theo vế những BĐT vừa thu được:
\(A\leq \sqrt{2}(x+y+z+3)+(2-\sqrt{2})\sqrt{3(x+y+z)}\)
\(\leq 6\sqrt{2}+(2-\sqrt{2}).3=6+3\sqrt{2}\)
Vậy \(A_{\max}=6+3\sqrt{2}\Leftrightarrow x=y=z=1\)
Áp dụng BĐT Bunhiacopxky:
(√1+x2+√2x)2≤(1+x2+2x)(1+1)(1+x2+2x)2≤(1+x2+2x)(1+1)
⇔√1+x2+√2x≤√2(x+1)⇔1+x2+2x≤2(x+1)
Hoàn toàn tt: {√1+y2+√2y≤√2(y+1)√1+z2+√2z≤√2(z+1){1+y2+2y≤2(y+1)1+z2+2z≤2(z+1)
Tiếp tục Bunhiacopxky:
(√x+√y+√z)2≤(x+y+z)(1+1+1)(x+y+z)2≤(x+y+z)(1+1+1)
⇒(2−√2)(√x+√y+√z)≤(2−√2)√3(x+y+z)⇒(2−2)(x+y+z)≤(2−2)3(x+y+z)
Cộng theo vế những BĐT vừa thu được:
A≤√2(x+y+z+3)+(2−√2)√3(x+y+z)A≤2(x+y+z+3)+(2−2)3(x+y+z)
≤6√2+(2−√2).3=6+3√2≤62+(2−2).3=6+32
Vậy Amax=6+3√2⇔x=y=z=1
