Đáp án: `A_{max}=\frac{\sqrt{183}}{2} ⇔ (x;y;z)=(\frac{17}{27};\frac{49}{36};\frac{217}{108})`
Giải thích các bước giải:
Áp dụng bất đẳng thức Bunyakovsky cho 2 bộ sô `(1;\sqrt{\frac{3}{2}};\sqrt{2}); (\sqrt{2x+1};\sqrt{2y+\frac{2}{3}};\sqrt{2z+\frac{1}{2}})` ta được:
`[1^2+(\sqrt{\frac{3}{2}})^2+(\sqrt{2})^2][(\sqrt{2x+1})^2+(\sqrt{2y+\frac{2}{3}})^2+(\sqrt{2z+\frac{1}{2}})^2]≥(1.\sqrt{2x+1}+\sqrt{\frac{3}{2}}.\sqrt{2y+\frac{2}{3}}+\sqrt{2}.\sqrt{2z+\frac{1}{2}})^2`
`⇔(1+\frac{3}{2}+2)(2x+1+2y+\frac{2}{3}+2z+\frac{1}{2})≥(\sqrt{2x+1}+\sqrt{3y+1}+\sqrt{4z+1})^2`
`⇔A^2≤\frac{9}{2}.[2(x+y+z)+\frac{13}{6}]=\frac{9}{2}.(2.4+\frac{13}{6})=\frac{183}{4}`
`⇔A≤\frac{\sqrt{183}}{2}`
Dấu bằng xảy ra
`⇔\frac{1}{\sqrt{2x+1}}=\frac{\sqrt{\frac{3}{2}}}{\sqrt{2y+\frac{2}{3}}}=\frac{\sqrt{2}}{\sqrt{2z+\frac{1}{2}}}`
`⇔\frac{1}{2x+1}=\frac{\frac{3}{2}}{2y+\frac{2}{3}}=\frac{2}{2z+\frac{1}{2}}`
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
`\frac{1}{2x+1}=\frac{\frac{3}{2}}{2y+\frac{2}{3}}=\frac{2}{2z+\frac{1}{2}}`
`=\frac{1+\frac{3}{2}+2}{2x+1+2y+\frac{2}{3}+2z+\frac{1}{2}}=\frac{\frac{9}{2}}{2.4+\frac{13}{6}}=\frac{27}{61}`
`\frac{1}{2x+1}=\frac{27}{61} ⇒ x=\frac{17}{27}`
`\frac{\frac{3}{2}}{2y+\frac{2}{3}}=\frac{27}{61} ⇒ y=\frac{49}{36}`
`\frac{2}{2z+\frac{1}{2}}=\frac{27}{61} ⇒ z=\frac{217}{108}`
