Cho a,b,c>0. Chứng minh rằng: \(\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge3\)
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H
10 tháng 7 2021
Áp dụng bất đẳng thức Svacxo ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{9}{a+2b}\)
Tương tự : \(\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{9}{b+2c};\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{9}{c+2a}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{a+2b}+\dfrac{3}{b+2c}+\dfrac{3}{c+2a}\)
Dấu = xảy ra khi a=b=c
10 tháng 7 2021
\(=>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{9}{a+2b}\)(BĐT Cauchy Schawarz)(1)
tương tự \(=>\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{9}{b+2c}\left(2\right)\)
\(=>\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{9}{c+2a}\left(3\right)\)
(1)(2)(3)
\(=>3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\)
\(=>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\left(dpcm\right)\)
11 tháng 9 2021
\(2\left(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}\right)\ge1+\dfrac{b}{b+1a}+\dfrac{c}{c+2b}+\dfrac{a}{a+2c}\)
\(\Leftrightarrow2\left(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}+\dfrac{a}{b+2a}+\dfrac{b}{c+2b}+\dfrac{c}{a+2c}\right)\ge1+\dfrac{b+2a}{b+2a}+\dfrac{c+2b}{c+2b}+\dfrac{a+2c}{a+2c}=1+1+1+1=4\)Thật vậy:
\(\dfrac{a}{b+2c}+\dfrac{a}{b+2a}+\dfrac{b}{c+2a}+\dfrac{b}{c+2b}+\dfrac{c}{a+2b}+\dfrac{c}{a+2c}=a\left(\dfrac{1}{b+2c}+\dfrac{1}{b+2a}\right)+b\left(\dfrac{1}{c+2a}+\dfrac{1}{c+2b}\right)+c\left(\dfrac{1}{a+2b}+\dfrac{1}{a+2c}\right)\)
\(\ge\dfrac{4a}{2\left(a+b+c\right)}+\dfrac{4b}{2\left(a+b+c\right)}+\dfrac{4c}{2\left(a+b+c\right)}=2\)
\(\Rightarrow VT\ge2.2=4\)
\(\RightarrowĐPCM\)
17 tháng 12 2022
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{2a+b}{2a-b}=\dfrac{2bk+b}{2bk-b}=\dfrac{2k+1}{2k-1}\)
\(\dfrac{2c+d}{2c-d}=\dfrac{2dk+d}{2dk-d}=\dfrac{2k+1}{2k-1}\)
=>\(\dfrac{2a+b}{2a-b}=\dfrac{2c+d}{2c-d}\)
b: \(\dfrac{2a+b}{a-2b}=\dfrac{2bk+b}{bk-2b}=\dfrac{2k+1}{k-2}\)
\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{2k+1}{k-2}\)
=>\(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)
14 tháng 5 2017
Áp dụng BĐT \(\left(x+y+z\right)\)\(\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)\(\ge9\) \(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\), ta có:
\(\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{2b}\ge\dfrac{9}{2a+4b}=\dfrac{9}{2\left(a+2b\right)}\)
\(\dfrac{1}{2b}+\dfrac{1}{2c}+\dfrac{1}{2c}\ge\dfrac{9}{2\left(b+2c\right)}\)
\(\dfrac{1}{2c}+\dfrac{1}{2a}+\dfrac{1}{2a}\ge\dfrac{9}{2\left(c+2a\right)}\)
Cộng từng vế ta được:
\(\dfrac{3}{2a}+\dfrac{3}{2b}+\dfrac{3}{2c}\ge\dfrac{9}{2\left(a+2b\right)}+\dfrac{9}{2\left(b+2c\right)}+\dfrac{9}{2\left(c+2a\right)}\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{9}{2}\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\)\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\left(đpcm\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 1 2022
\(\dfrac{a^3}{\left(a+2b\right)\left(b+2c\right)}+\dfrac{a+2b}{27}+\dfrac{b+2c}{27}\ge3\sqrt[3]{\dfrac{a^3\left(a+2b\right)\left(b+2c\right)}{27^2.\left(a+2b\right)\left(b+2c\right)}}=\dfrac{a}{3}\)
Tương tự:
\(\dfrac{b^3}{\left(b+2c\right)\left(c+2a\right)}+\dfrac{b+2c}{27}+\dfrac{c+2a}{27}\ge\dfrac{b}{3}\)
\(\dfrac{c^3}{\left(c+2a\right)\left(a+2b\right)}+\dfrac{c+2a}{27}+\dfrac{a+2b}{27}\ge\dfrac{c}{3}\)
Cộng vế:
\(VT+\dfrac{2\left(a+b+c\right)}{9}\ge\dfrac{a+b+c}{3}\)
\(\Rightarrow VT\ge\dfrac{a+b+c}{9}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
20 tháng 5 2023
\(A=\dfrac{a^2}{ab+2ac}+\dfrac{b^2}{bc+2ab}+\dfrac{c^2}{ca+2bc}>=\dfrac{\left(a+b+c\right)^2}{3\left(ab+bc+ac\right)}\)>=1
2 tháng 7 2021
\(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}=\dfrac{a^2}{ab+2ac}+\dfrac{b^2}{bc+2ab}+\dfrac{c^2}{ac+2bc}\)
áp dụng BDT CAUCHY SCHAWRZ
\(=>\dfrac{a^2}{ab+2ac}+\dfrac{b^2}{bc+2ab}+\dfrac{c^2}{ac+2bc}\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ac+2ac+2ab+2bc}\)
\(=\dfrac{\left(a+b+c\right)^2}{3\left(ab+bc+ac\right)}\ge\dfrac{3\left(ab+bc+ac\right)}{3\left(ab+bc+ac\right)}=1\)
PN
5 tháng 7 2021
cái chỗ bđt cauchy là bđt gì bạn có thể ghi cụ thể nó ra được ko ạ
7 tháng 9 2021
\(\dfrac{bc}{a+b+c+a}\le\dfrac{bc}{4}\cdot\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\ \dfrac{ac}{b+c+a+b}\le\dfrac{ac}{4}\cdot\left(\dfrac{1}{b+c}+\dfrac{1}{a+b}\right)\\ \dfrac{ab}{a+c+b+c}\le\dfrac{ab}{4}\cdot\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\\ \Leftrightarrow VT\le\dfrac{1}{a+b}\left(\dfrac{bc}{4}+\dfrac{ac}{4}\right)+\dfrac{1}{a+c}\left(\dfrac{bc}{4}+\dfrac{ab}{4}\right)+\dfrac{1}{b+c}\left(\dfrac{ac}{4}+\dfrac{ab}{4}\right)\\ =\dfrac{1}{a+b}\cdot\dfrac{c\left(a+b\right)}{4}+\dfrac{1}{a+c}\cdot\dfrac{b\left(a+c\right)}{4}+\dfrac{1}{b+c}\cdot\dfrac{a\left(b+c\right)}{4}\\ =\dfrac{c}{4}+\dfrac{b}{4}+\dfrac{a}{4}\\ =\dfrac{a+b+c}{4}\left(đfcm\right)\)
