cho số dương a,b,c. Tìm GTLN : \(\dfrac{ab}{a^2+ab+bc}+\dfrac{bc}{b^2+bc+ca}+\dfrac{ca}{c^2+ca+ab}\)
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8 tháng 5 2022
\(M=\dfrac{1}{\dfrac{c}{a}+\dfrac{2a}{b}+3}+\dfrac{1}{\dfrac{a}{b}+\dfrac{2b}{c}+3}+\dfrac{1}{\dfrac{b}{c}+\dfrac{2c}{a}+3}\)
\(đặt\left(\dfrac{a}{b};\dfrac{b}{c};\dfrac{c}{a}\right)=\left(x;y;z\right)\Rightarrow xyz=1\left(x;y;z>0\right)\)
\(M=\dfrac{1}{z+2x+3}+\dfrac{1}{x+2y+3}+\dfrac{1}{y+2z+3}\)
\(ta\) \(đi\) \(cminh:A\le\dfrac{1}{2}\)
có:
\(\dfrac{1}{z+2x+3}\le\dfrac{1}{6}\Leftrightarrow z+2x+3\ge6\Leftrightarrow2x+z\ge3\)
\(\dfrac{1}{x+2y+3}\le\dfrac{1}{6}\Leftrightarrow x+2y\ge3\)
\(\dfrac{1}{y+2z+3}\le\dfrac{1}{6}\Rightarrow y+2z\ge3\)
\(cộng\) \(vế\Rightarrow2x+z+2y+x+2z+y\ge9\Leftrightarrow x+y+z\ge3\left(đúng\right)\)
\(do:x+y+z\ge3\sqrt[3]{xyz}=3\)
\(\Rightarrow A\le\dfrac{1}{2}dấu"="\Leftrightarrow x=y=z=1\Rightarrow a=b=c\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 5 2021
Đề bài có nhầm lẫn gì ko nhỉ?
\(T=\dfrac{ab}{a^2+b^2+ab}+\dfrac{bc}{b^2+c^2+2bc}+\dfrac{ca}{c^2+a^2+ca}\le\dfrac{ab}{2ab+ab}+\dfrac{bc}{2bc+bc}+\dfrac{ca}{2ca+ca}=1\)
AH
Akai Haruma
Giáo viên
28 tháng 10 2021
Lời giải:
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\Rightarrow \frac{abc}{c(a+b)}=\frac{abc}{a(b+c)}=\frac{bca}{b(c+a)}\)
\(\Leftrightarrow c(a+b)=a(b+c)=b(c+a)\)
\(\Leftrightarrow ac+bc=ab+ac=bc+ab\Leftrightarrow ab=bc=ac\)
\(\Rightarrow a=b=c\) (do $a,b,c>0$)
$\Rightarrow M=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1$
30 tháng 12 2021
\(\dfrac{ab}{\sqrt{ab+2c}}=\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}=ab\cdot\sqrt{\dfrac{1}{a+b}\cdot\dfrac{1}{b+c}}\le ab\cdot\dfrac{1}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)=\dfrac{1}{2}\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}\right)\)
CMTT: \(\dfrac{bc}{\sqrt{bc+2a}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right);\dfrac{ac}{\sqrt{ac+2b}}\le\dfrac{1}{2}\left(\dfrac{ac}{b+c}+\dfrac{ac}{b+a}\right)\)
\(\Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{ab}{c+a}+\dfrac{ab}{c+b}+\dfrac{bc}{b+a}+\dfrac{bc}{c+a}+\dfrac{ac}{b+c}+\dfrac{ac}{b+c}\right)\\ \Leftrightarrow P\le\dfrac{1}{2}\left[\dfrac{b\left(a+c\right)}{a+c}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{c\left(a+b\right)}{a+b}\right]=\dfrac{1}{2}\left(a+b+c\right)=1\)
Dấu \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2021
Cần điều kiện a;b;c dương
\(\dfrac{bc}{\sqrt{a.1+bc}}=\dfrac{bc}{\sqrt{a\left(a+b+c\right)+bc}}=\dfrac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right)\)
Tương tự: \(\dfrac{ca}{\sqrt{b+ca}}\le\dfrac{1}{2}\left(\dfrac{ca}{a+b}+\dfrac{ca}{b+c}\right)\) ; \(\dfrac{ab}{\sqrt{c+ab}}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)
Cộng vế với vế:
\(A\le\dfrac{1}{2}\left(\dfrac{bc+ca}{a+b}+\dfrac{bc+ab}{a+c}+\dfrac{ca+ab}{b+c}\right)=\dfrac{1}{2}\left(a+b+c\right)=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
\(VT=\dfrac{1}{\dfrac{a}{b}+\dfrac{c}{a}+1}+\dfrac{1}{\dfrac{b}{c}+\dfrac{a}{b}+1}+\dfrac{1}{\dfrac{c}{a}+\dfrac{b}{c}+1}\)
\(\left(\dfrac{a}{b},\dfrac{b}{c},\dfrac{c}{a}\right)\rightarrow\left(x^3,y^3,z^3\right)\)\(\Rightarrow xyz=1\).
\(VT=\sum\dfrac{1}{x^3+y^3+1}\le\sum\dfrac{1}{xy\left(x+y\right)+xyz}=\sum\dfrac{z}{x+y+z}=1\)
Dấu = xảy ra khi x=y=z=1 hay a=b=c
Hay quá bạn ơi tks