Cho a,b,c là 3 số thực dương tùy ý Chứn minh rằng
\(\dfrac{a}{\sqrt{ab+b^2}}+\dfrac{b}{\sqrt{bc+b^2}}+\dfrac{c}{\sqrt{ac+a^2}}\ge\dfrac{3\sqrt{2}}{2}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
23 tháng 9 2021
Áp dụng BĐT Cosi:
\(\dfrac{a}{\sqrt{b^2+ab}}=\dfrac{a\sqrt{2}}{\sqrt{2\left(b^2+ab\right)}}=\dfrac{a\sqrt{2}}{\sqrt{2b\left(a+b\right)}}\ge\dfrac{a\sqrt{2}}{\dfrac{2b+a+b}{2}}=\dfrac{2\sqrt{2}a}{a+3b}\)
Cmtt: \(\dfrac{b}{\sqrt{c^2+bc}}\ge\dfrac{2\sqrt{2}b}{b+3c};\dfrac{c}{\sqrt{a^2+ca}}\ge\dfrac{2\sqrt{2}c}{c+3a}\)
\(\Leftrightarrow P\ge2\sqrt{2}\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)\\ \Leftrightarrow\dfrac{P}{\sqrt{2}}\ge2\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)\\ \Leftrightarrow\dfrac{P}{\sqrt{2}}\ge\dfrac{2\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\dfrac{1}{3}\left(a+b+c\right)^2}\\ \Leftrightarrow\dfrac{P}{\sqrt{2}}\ge\dfrac{2}{\dfrac{4}{3}}=\dfrac{3}{2}\\ \Leftrightarrow P\ge\dfrac{3\sqrt{2}}{2}\)
Dấu \("="\Leftrightarrow a=b=c\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 4 2022
Đề bài sai
Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)
KK
7 tháng 3 2017
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
Chứng minh rằng \(\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)
\(\Leftrightarrow18\ge3\left(3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}\right)\)
\(\Leftrightarrow18\ge9+3\sqrt[3]{bc}+3\sqrt[3]{ca}+3\sqrt[3]{ab}\)
\(\Leftrightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)
Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}a+b+1\ge3\sqrt[3]{ab}\\b+c+1\ge3\sqrt[3]{bc}\\c+a+1\ge3\sqrt[3]{ca}\end{matrix}\right.\)
\(\Rightarrow2\left(a+b+c\right)+3\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)
\(\Rightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\) ( đpcm )
Vì \(\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)
Mà \(\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)( đpcm )
8 tháng 3 2017
Áp dụng BĐT AM-GM và Cauchy-Schwarz ta có:
\(\sum\frac{a^2}{a+\sqrt[3]{bc}}\geq\sum\frac{a^2}{a+\frac{b+c+1}{3}}=\sum\frac{9a^2}{3(3a+b+c)+a+b+c}\)
\(=\sum\frac{9a^2}{10a+4b+4c}\geq\frac{9(a+b+c)^2}{(10a+4b+4c)}=\frac{9(a+b+c)^2}{18(a+b+c)}=\frac{3}{2}\)
\(\sum\dfrac{a}{\sqrt{ab+b^2}}=\sum\dfrac{a\sqrt{2}}{\sqrt{2b\left(a+b\right)}}\ge\sum\dfrac{2\sqrt{2}a}{2b+a+b}=2\sqrt{2}\sum\dfrac{a}{a+3b}\)
\(=2\sqrt{2}\sum\dfrac{a^2}{a^2+3ab}\ge\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\)
\(=\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{\left(a+b+c\right)^2+ab+bc+ca}\ge\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\dfrac{1}{3}\left(a+b+c\right)^2}=\dfrac{3\sqrt{2}}{2}\)