cho a,b,c>0.cmr1/(2a+b)+1/(2b+c)+1/(2c+a)>=3/(a+b+c)
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NY
26 tháng 8 2020
Ta có :
\(a^2b+b^2c+c^2a\ge\frac{9a^2b^2c^2}{1+2a^2b^2c^2}\)
\(\Leftrightarrow\left(a^2b+b^2c+c^2a\right)\left(1+2a^2b^2c^2\right)\ge9a^2b^2c^2\)
\(\Leftrightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^{3v}+2a^3b^2c^4\ge3a^2b^2c^2\left(a+b+c\right)\)(*)
Áp dụng BĐT AM-GM ta có:
\(a^2b+a^4b^3c^2+a^3b^2c^4\ge3\sqrt[3]{a^9b^6c^6}=3a^3b^2c^2\)
\(b^2c+a^2b^4c^3+a^4b^3c^2\ge3a^2b^3c^2\)
\(c^2a+a^3b^2c^4+a^2b^4c^4\ge3a^2b^2c^3\)
Cộng theo vế
\(\Rightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^3+2a^3b^2c^4\ge3a^2b^2c^2\left(a+b+c\right)\)
Vậy $(*)$ đúng
Do đó ta có đpcm
#Cừu
DX
0
31 tháng 1 2019
\(\frac{3}{a+2b}=\frac{1}{3}.\frac{9}{a+b+b}\le\frac{1}{3}.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\right)\)
Tương tự:\(\frac{3}{b+2c}\le\frac{1}{3}\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\frac{3}{c+2a}\le\frac{1}{3}\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{a}\right)\)
Cộng theo vế ta được:
\(\frac{3}{a+2b}+\frac{3}{b+2c}+\frac{3}{c+2a}\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}+\frac{1}{a}\right)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
HP
30 tháng 4 2017
Có a,b,c>0;a+b>c,b+c>a,c+a>b
=>a+b-c>0,b+c-a>0,c+a-b>0
=>c2(a+b-c)>0,a2(b+c-a)>0,b2(c+a-b)>0
=>c2(a+b-c)+a2(b+c-a)+b2(c+a-b)>0
=>(đẳng thức đề bài) > 0
HD
1
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 3 2019
Với các số dương, áp dụng BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)
\(\frac{1}{a+b}+\frac{1}{b+c}\ge\frac{4}{a+2b+c}\) ; \(\frac{1}{a+b}+\frac{1}{a+c}\ge\frac{4}{2a+b+c}\); \(\frac{1}{b+c}+\frac{1}{a+c}\ge\frac{4}{a+b+2c}\)
Cộng vế với vế:
\(2\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\ge4\left(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\right)\)
\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\ge2\left(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\right)\)
Dấu "=" xảy ra khi \(a=b=c\)
DA
1
G
11 tháng 3 2019
Áp dụng bđt Cauchy-Schwarz
\(\frac{1}{2a+b+c}=\frac{1}{\left(a+b\right)+\left(a+c\right)}\le\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)
\(\frac{1}{a+2b+c}=\frac{1}{\left(a+b\right)+\left(b+c\right)}\le\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)\)
\(\frac{1}{a+b+2c}=\frac{1}{\left(a+c\right)+\left(b+c\right)}\le\frac{1}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)
Cộng theo vế =>đpcm
HN
0
18 tháng 11 2019
Bài 2:
\(\frac{1}{\sqrt[3]{81}}\cdot P=\frac{1}{\sqrt[3]{9\cdot9\cdot\left(a+2b\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(b+2c\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(c+2a\right)}}\)
\(\ge\frac{3}{a+2b+9+9}+\frac{3}{b+2c+9+9}+\frac{3}{c+2a+9+9}\ge3\left(\frac{9}{3a+3b+3c+54}\right)=\frac{1}{3}\)
\(\Rightarrow P\ge\sqrt[3]{3}\)
Dấu bằng xẩy ra khi a=b=c=3
18 tháng 11 2019
Bài 1:
\(ab+bc+ca=5abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=5\)
Theo bđt côsi-shaw ta luôn có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge\frac{25}{x+y+z+t+k}\)(x=y=z=t=k>0 ) (*)
\(\Leftrightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)
Áp dụng bđt AM-GM ta có:
\(\hept{\begin{cases}x+y+z+t+k\ge5\sqrt[5]{xyztk}\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge5\sqrt[5]{\frac{1}{xyztk}}\end{cases}}\)
\(\Rightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)
\(\Rightarrow\)(*) luôn đúng
Từ (*) \(\Rightarrow\frac{1}{25}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\le\frac{1}{x+y+z+t+k}\)
Ta có: \(P=\frac{1}{2a+2b+c}+\frac{1}{a+2b+2c}+\frac{1}{2a+b+2c}\)
Mà \(\frac{1}{2a+2b+c}=\frac{1}{a+a+b+b+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\frac{1}{a+2b+2c}=\frac{1}{a+b+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\frac{1}{2a+b+2c}=\frac{1}{a+a+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\Rightarrow P\le\frac{1}{25}\left[5.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]=1\)
\(\Rightarrow P\le1\left(đpcm\right)\)Dấu"="xảy ra khi a=b=c\(=\frac{3}{5}\)
Áp dụng bất đẳng thức Cauchy-Schwartz, ta có: \(\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\ge\frac{\left(1+1+1\right)^2}{2a+b+2b+c+2c+a}=\frac{9}{3\left(a+b+c\right)}=\frac{3}{a+b+c}\)
Dấu "=" xảy ra khi: \(\frac{1}{2a+b}=\frac{1}{2b+c}=\frac{1}{2c+a}\Leftrightarrow2a+b=2b+c=2c+a\)