Cho a+b+c=2021. tìm gtln \(\frac{ab}{\sqrt{ab+2021c}}+\frac{bc}{\sqrt{bc+2021a}}+\frac{ca}{\sqrt{ca+2021b}}\)
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15 tháng 7 2020
\(\sum\frac{a}{a+\sqrt{2021a+bc}}=\sum\frac{a}{a+\sqrt{a\left(a+b+c\right)+bc}}=\sum\frac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\le_{C-S}\sum\frac{a}{a+\sqrt{\left(\sqrt{ab}+\sqrt{ac}\right)^2}}=\sum\frac{a}{a+\sqrt{ab}+\sqrt{ac}}=\sum\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\)
C
Cho a+b+c=1. Tìm GTLN của
\(A=\sqrt{\frac{ab}{c+ab}}+\sqrt{\frac{bc}{a+bc}}+\sqrt{\frac{ca}{b+ca}}\)
1
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 2 2021
Cần thêm điều kiện a;b;c dương
\(\sqrt{\dfrac{ab}{c+ab}}=\sqrt{\dfrac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\)
Tương tự: \(\sqrt{\dfrac{bc}{a+bc}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}\right)\) ; \(\sqrt{\dfrac{ca}{b+ac}}\le\dfrac{1}{2}\left(\dfrac{c}{b+c}+\dfrac{a}{a+b}\right)\)
Cộng vế với vế:
\(A\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{a}{a+c}+\dfrac{c}{a+c}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
TN
19 tháng 5 2017
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(3P=\frac{3\sqrt{ab}}{c+3\sqrt{bc}}+\frac{3\sqrt{bc}}{a+3\sqrt{bc}}+\frac{3\sqrt{ca}}{b+3\sqrt{ca}}\)
\(=3-\left(\frac{a}{a+3\sqrt{bc}}+\frac{b}{b+3\sqrt{ca}}+\frac{c}{c+3\sqrt{ab}}\right)\)
\(\le3-\left[\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}\right]\)
\(\le3-\left[\frac{\left(a+b+c\right)^2}{\left(a^2+b^2+c^2\right)+3\left(ab+bc+ca\right)}\right]\)
\(\le3-\left[\frac{\left(a+b+c\right)^2}{\left(a^2+b^2+c^2\right)+\frac{\left(a+b+c\right)^2}{3}}\right]=3-\frac{9}{4}=\frac{3}{4}\)
Đẳng thức xảy ra khi \(a=b=c\)
20 tháng 5 2017
Bạn Thắng Nguyễn chạy mà bị kiệt sức ở giai đoạn cuối để mình chạy tiếp sức phần còn lại nhé.
Từ \(3-\left[\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\right]\)
\(\le3-\left[\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\frac{\left(a+b+c\right)^2}{3}}\right]=3-\frac{3}{4}=\frac{9}{4}\)
\(\Rightarrow P\le\frac{\frac{9}{4}}{3}=\frac{3}{4}\)
9 tháng 8 2020
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
20 tháng 8 2017
\(P=\frac{ab}{\sqrt{\left(c+a\right)\left(b+c\right)}}+\frac{bc}{\sqrt{\left(c+a\right)\left(a+b\right)}}+\frac{ca}{\sqrt{\left(b+c\right)\left(a+b\right)}}\)
thử dùng cô si đi
8 tháng 6 2016
Ta có: \(\frac{bc}{\sqrt{a+bc}}=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{\frac{bc}{a+b}+\frac{bc}{a+c}}{2}\)
\(\frac{ca}{\sqrt{b+ac}}=\frac{ca}{\sqrt{\left(a+b\right)\left(b+c\right)}}\le\frac{\frac{ca}{a+b}+\frac{ca}{b+c}}{2}\)
\(\frac{ab}{\sqrt{c+ab}}=\frac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\frac{\frac{ab}{a+c}+\frac{ab}{b+c}}{2}\)
Cộng 3 vế ta được: \(P\le\frac{\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ca}{a+b}+\frac{ca}{b+c}+\frac{ab}{a+c}+\frac{ab}{b+c}}{2}\)
\(=\frac{\frac{c\left(a+b\right)}{a+b}+\frac{b\left(a+c\right)}{a+c}+\frac{a\left(b+c\right)}{b+c}}{2}=\frac{a+b+c}{2}=\frac{1}{2}\)
Vậy MinP = 1/2
8 tháng 6 2016
\(\frac{bc}{\sqrt{a+bc}}=\frac{bc}{\sqrt{a.1+bc}}=\frac{bc}{\sqrt{a\left(a+b+c\right)+bc}}=\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{bc}{a+b}+\frac{bc}{a+c}\)
\(\frac{ab}{\sqrt{ab+2021c}}+\frac{bc}{\sqrt{bc+2021a}}+\frac{ca}{\sqrt{ca+2021b}}\)
\(=\frac{ab}{\sqrt{ab+\left(a+b+c\right)c}}+\frac{bc}{\sqrt{bc+\left(a+b+c\right)a}}+\frac{ca}{\sqrt{ca+\left(a+b+c\right)b}}\)
\(=\frac{ab}{\sqrt{\left(c+a\right)\left(c+b\right)}}+\frac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{ca}{\sqrt{\left(b+a\right)\left(b+c\right)}}\)
\(\le\frac{1}{2}\left(\frac{ab}{c+a}+\frac{ab}{c+b}+\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ca}{b+a}+\frac{ca}{b+c}\right)\)
\(=\frac{1}{2}\left[\left(\frac{ab}{c+a}+\frac{bc}{a+c}\right)+\left(\frac{ab}{c+b}+\frac{ca}{b+c}\right)+\left(\frac{bc}{a+b}+\frac{ca}{a+b}\right)\right]\)
\(=\frac{1}{2}\left(a+b+c\right)=\frac{2012}{2}=1006\)