Toán Lớp 10: Tìm GTLN của biểu thức M = $\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}$ với a,b,c>0, a+b+c=1.
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 10: Tìm GTLN của biểu thức M = $\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}$ với a,b,c>0, a+b+c=1.
Comments ( 1 )
Cách 1: Dùng BĐT Bunhiacopxki
Áp dụng BĐT Bunhiacopxki, ta có:
M^2 = (\sqrt{a + 1} + \sqrt{b + 1} + \sqrt{c + 1})^2
= (1.\sqrt{a + 1} + 1. \sqrt{b + 1} + 1. \sqrt{c + 1})^2 ≤ (1^2 + 1^2 + 1^2)[(\sqrt{a + 1})^2 + (\sqrt{b + 1})^2 + (\sqrt{c + 1})^2] = 3(a + 1 + b + 1 + c + 1) = 3(1 + 3) = 12
$\\$
Ta được: M^2 ≤ 12
⇔ M ≤ 2\sqrt{3}
Vậy GTLN của M là: 2\sqrt{3} khi: a = b = c = 1/3
Cách 2: Dùng BĐT Cô – si
Áp dụng BĐT Cô – si, ta có:
$\dfrac{2}{\sqrt{3}}$.\sqrt{a + 1} <= \frac{(2/\sqrt{3})^2 + (\sqrt{a + 1})^2}{2} = $\dfrac{\dfrac{4}{3} + a + 1}{2}$ = $\dfrac{\dfrac{7}{3} + a}{2}$
$\\$
$\dfrac{2}{\sqrt{3}}$.\sqrt{b + 1} <= \frac{(2/\sqrt{3})^2 + (\sqrt{b + 1})^2}{2} = $\dfrac{\dfrac{4}{3} + b + 1}{2}$ = $\dfrac{\dfrac{7}{3} + b}{2}$
$\\$
$\dfrac{2}{\sqrt{3}}$.\sqrt{c + 1} <= \frac{(2/\sqrt{3})^2 + (\sqrt{c + 1})^2}{2} = $\dfrac{\dfrac{4}{3} + c + 1}{2}$ = $\dfrac{\dfrac{7}{3} + c }{2}$
⇒ $\dfrac{2}{\sqrt{3}}$.\sqrt{a + 1} + $\dfrac{2}{\sqrt{3}}$.\sqrt{b + 1} + $\dfrac{2}{\sqrt{3}}$.\sqrt{c + 1} ≤ $\dfrac{\dfrac{7}{3} + a}{2}$ + $\dfrac{\dfrac{7}{3} + b}{2}$ + $\dfrac{\dfrac{7}{3} + c}{2}$
⇔ $\dfrac{2}{\sqrt{3}}$.(\sqrt{a + 1} + \sqrt{b + 1} + \sqrt{c + 1}) ≤ $\dfrac{\dfrac{7}{3} + a}{2}$ + $\dfrac{\dfrac{7}{3} + b}{2}$ + $\dfrac{\dfrac{7}{3} + c}{2}$
$\\$
⇔ $\dfrac{2}{\sqrt{3}}$.M ≤ $\dfrac{\dfrac{7}{3} + a}{2}$ + $\dfrac{\dfrac{7}{3} + b}{2}$ + $\dfrac{\dfrac{7}{3} + c}{2}$ = $\dfrac{\dfrac{7}{3} + a + \dfrac{7}{3} + b + \dfrac{7}{3} + c}{2}$ = $\dfrac{\dfrac{7}{3} + \dfrac{7}{3} + \dfrac{7}{3} + 1}{2}$ = 4
$\\$
⇔ M ≤ $\dfrac{4}{\dfrac{2}{\sqrt{3}}}$ = 2\sqrt{3}
Vậy GTLN của M là: 2\sqrt{3} khi: a = b = c = 1/3