Chủ đề này có 2 trả lời

[diemdaotran]

a,b,c là các số thực dương thỏa mãn $\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}=2$ 

Tìm Max P=$\frac{1}{\sqrt{a+3b}}+\frac{1}{\sqrt{b+3c}}+\frac{1}{\sqrt{c+3a}}$

Bài viết đã được chỉnh sửa nội dung bởi tienduc: 07-07-2017 - 08:14

[HoangKhanh2002]

a,b,c là các số thực dương thỏa mãn $\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}=2$ 

Tìm Max P=$\frac{1}{\sqrt{a+3b}}+\frac{1}{\sqrt{b+3c}}+\frac{1}{\sqrt{c+3a}}$

Áp dụng $AM-GM$ ta có:

$P=\dfrac{1}{\sqrt{a+3b}}+\dfrac{1}{\sqrt{b+3c}}+\dfrac{1}{\sqrt{c+3a}}\leqslant \dfrac{1}{\sqrt{4\sqrt[4]{ab^3}}}+ \dfrac{1}{\sqrt{4\sqrt[4]{bc^3}}}+ \dfrac{1}{\sqrt{4\sqrt[4]{ca^3}}}\\ =\dfrac{1}{2}\left ( \dfrac{1}{\sqrt[8]{ab^3}}+ \dfrac{1}{\sqrt[8]{bc^3}}+ \dfrac{1}{\sqrt[8]{ca^3}}\right )\leqslant \dfrac{1}{8}\left ( \dfrac{1}{\sqrt{a}}+\dfrac{3}{\sqrt{b}}+\dfrac{1}{\sqrt{b}}+\dfrac{3}{\sqrt{c}}+\dfrac{1}{\sqrt{c}}+\dfrac{3}{\sqrt{a}} \right ) \\=\dfrac{1}{2}\left ( \dfrac{1}{\sqrt{a}} +\dfrac{1}{\sqrt{b}} +\dfrac{1}{\sqrt{c}} \right )=1$

Dấu "=" xảy ra: $\iff a=b=c=\dfrac{9}{4} \hspace{0,5cm} \square$

[diemdaotran]

Áp dụng $AM-GM$ ta có:

$P=\dfrac{1}{\sqrt{a+3b}}+\dfrac{1}{\sqrt{b+3c}}+\dfrac{1}{\sqrt{c+3a}}\leqslant \dfrac{1}{\sqrt{4\sqrt[4]{ab^3}}}+ \dfrac{1}{\sqrt{4\sqrt[4]{bc^3}}}+ \dfrac{1}{\sqrt{4\sqrt[4]{ca^3}}}\\ =\dfrac{1}{2}\left ( \dfrac{1}{\sqrt[8]{ab^3}}+ \dfrac{1}{\sqrt[8]{bc^3}}+ \dfrac{1}{\sqrt[8]{ca^3}}\right )\leqslant \dfrac{1}{8}\left ( \dfrac{1}{\sqrt{a}}+\dfrac{3}{\sqrt{b}}+\dfrac{1}{\sqrt{b}}+\dfrac{3}{\sqrt{c}}+\dfrac{1}{\sqrt{c}}+\dfrac{3}{\sqrt{a}} \right ) \\=\dfrac{1}{2}\left ( \dfrac{1}{\sqrt{a}} +\dfrac{1}{\sqrt{b}} +\dfrac{1}{\sqrt{c}} \right )=1$

Dấu "=" xảy ra: $\iff a=b=c=\dfrac{9}{4} \hspace{0,5cm} \square$

 

ADBDT Bunhiacopxki ta duoc:

$(\frac{9}{4}+\frac{27}{4})(a+3b)\geq (\frac{3}{2}\sqrt{a}+\frac{9}{2}\sqrt{b})^2 \Rightarrow \sqrt{a+3b}\geq (\frac{3}{2}\sqrt{a}+\frac{9}{2}\sqrt{b}):3\Rightarrow \frac{1}{\sqrt{a+3b}}\leq 1:(\frac{1}{2}\sqrt{a}+\frac{3}{2}\sqrt{b}) \Rightarrow \frac{1}{\sqrt{a+3b}}\leq \frac{2}{\sqrt{a}+3\sqrt{b}}$

Tuong tu cong ve ta duoc$\frac{1}{\sqrt{a+3b}}+\frac{1}{\sqrt{b+3c}}+\frac{1}{\sqrt{c+3a}}\leq \frac{2}{\sqrt{a}+3\sqrt{b}}+\frac{2}{\sqrt{b}+3\sqrt{c}}+\frac{2}{\sqrt{c}+3\sqrt{a}}$ (1)

Lai co $\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{b}}\geq \frac{16}{\sqrt{a}+3\sqrt{b}}$

Tương tự công về lại được $\frac{4}{\sqrt{a}}+\frac{4}{\sqrt{b}}+\frac{4}{\sqrt{c}}\geq \frac{16}{\sqrt{a}+3\sqrt{b}}+\frac{16}{\sqrt{b}+3\sqrt{c}}+\frac{16}{\sqrt{c}+3\sqrt{a}}\Rightarrow \frac{1}{2}(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})\geq \frac{2}{\sqrt{a}+3\sqrt{b}}+\frac{2}{\sqrt{b}+3\sqrt{c}}+\frac{2}{\sqrt{c}+3\sqrt{a}}$  (2)

Tu (1) va (2) duoc $\frac{1}{2}(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})\geq \frac{1}{\sqrt{a+3b}}+\frac{1}{\sqrt{b+3c}}+\frac{1}{\sqrt{c+3a}}\Rightarrow 1\geq \frac{1}{\sqrt{a+3b}}+\frac{1}{\sqrt{b+3c}}+\frac{1}{\sqrt{c+3a}}$

Bài viết đã được chỉnh sửa nội dung bởi diemdaotran: 07-07-2017 - 07:24