Tìm GTLN của biểu thức M = $\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}$ với a,b,c>0, a+b+c=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`