Cho a+b+c=4,5. CHứng minh \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}>1\)
K
Khách
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KK
5 tháng 2 2017
Áp dụng bất đẳng thức \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\left\{\begin{matrix}\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\\\frac{1}{b+c}\le\frac{1}{4}\left(\frac{1}{b}+\frac{1}{c}\right)\\\frac{1}{a+c}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{c}\right)\end{matrix}\right.\)
Cộng theo từng vế:
\(\Rightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\le\frac{1}{4}\left(\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\right)\)
\(\Rightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\le\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) ( đpcm )
5 tháng 2 2017
Với a , b , c > 0
Ta có: \(a^2-2ab+b^2\ge0\)
\(\Rightarrow\) \(a^2+2ab+b^2\ge4ab\)
\(\Rightarrow\) \(\left(a+b\right)^2\ge4ab\)
\(\Rightarrow\) \(\frac{a+b}{4ab}\ge\frac{1}{a+b}\)
\(\Rightarrow\) \(\frac{1}{a+b}\le\frac{1}{4b}+\frac{1}{4a}\)
\(\Rightarrow\) \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)(1)
Chứng minh tương tự ta cũng có được:
\(\frac{1}{b+c}\le\frac{1}{4}\left(\frac{1}{b}+\frac{1}{c}\right)\) (2)
và \(\frac{1}{a+c}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{c}\right)\) (3)
Cộng (1), (2), (3) vế theo vế ta được:
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{c}\right)\)
\(\Rightarrow\) \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\le\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)( ĐPCM)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 11 2019
a/ Đề sai, đề đúng phải là \(p=\frac{a+b+c}{2}\)
b/ \(\Leftrightarrow\frac{2}{2+a^2b}+\frac{2}{2+b^2c}+\frac{2}{2+c^2a}\ge2\)
\(VT=1-\frac{a^2b}{1+1+a^2b}+1-\frac{b^2c}{1+1+b^2c}+1-\frac{c^2a}{1+1+c^2a}\)
\(VT\ge3-\left(\frac{a^2b}{3\sqrt[3]{a^2b}}+\frac{b^2c}{3\sqrt[3]{b^2c}}+\frac{c^2a}{3\sqrt[3]{c^2a}}\right)\)
\(VT\ge3-\frac{1}{9}\left(3\sqrt[3]{a^2.ab.ab}+3\sqrt[3]{b^2.bc.bc}+3\sqrt[3]{c^2.ca.ca}\right)\)
\(VT\ge3-\frac{1}{9}\left(a^2+2ab+b^2+2bc+c^2+2ca\right)\)
\(VT\ge3-\frac{1}{9}\left(a+b+c\right)^2=2\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
HN
Cho a>1,b>1,c>1. Chứng minh : \(\frac{a}{\sqrt{a-1}}+\frac{b}{\sqrt{b-1}}+\frac{c}{\sqrt{c-1}}\ge6\)
1
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 7 2020
\(VT=\frac{a}{\sqrt{a-1}}+\frac{b}{\sqrt{b-1}}+\frac{c}{\sqrt{c-1}}=\frac{a-1+1}{\sqrt{a-1}}+\frac{b-1+1}{\sqrt{b-1}}+\frac{c-1+1}{\sqrt{c-1}}\)
\(VT\ge\frac{2\sqrt{a-1}}{\sqrt{a-1}}+\frac{2\sqrt{b-1}}{\sqrt{b-1}}+\frac{2\sqrt{c-1}}{\sqrt{c-1}}=6\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=2\)
20 tháng 3 2020
BĐT phụ:\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Leftrightarrow\left(x-y\right)^2\ge0\left(true\right)\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{4}{a+b}+\frac{1}{c}\ge\frac{9}{a+b+c}\) ( đpcm )
Vậy.......
T
13 tháng 2 2020
Mấy cái dấu "=" anh tự xét.
Áp dụng BĐT AM-GM: \(VT=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=\frac{3}{\sqrt[3]{abc}}\ge\frac{3}{\frac{a+b+c}{3}}=\frac{9}{a+b+c}\)
a) Áp dụng: \(VT\ge\frac{\left(a+b+c\right)^2}{3}.\frac{9}{2\left(a+b+c\right)}=\frac{3}{2}\left(a+b+c\right)\)
b) \(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{x+y+z+3}=\frac{3}{4}\)
21 tháng 7 2020
a) Chứng minh được BĐT \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)(*)
Dấu "=" xảy ra <=> a=b
Áp dụng BĐT (*) vào bài toán ta có:
\(\hept{\begin{cases}\frac{1}{2x+y+z}=\frac{1}{x+y+x+y}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\\\frac{1}{x+2y+z}=\frac{1}{x+y+y+z}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\\\frac{1}{x+y+2z}=\frac{1}{x+y+z+z}\le\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\end{cases}}\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\cdot2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
Tiếp tục áp dụng BĐT (*) ta có:
\(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right);\frac{1}{y+z}\le\frac{1}{4}\left(\frac{1}{y}+\frac{1}{z}\right);\frac{1}{z+x}\le\frac{1}{4}\left(\frac{1}{z}+\frac{1}{x}\right)\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\cdot2\cdot\frac{1}{4}\cdot2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)
\(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\)
Dấu "=" xảy ra <=> \(x=y=z=\frac{3}{4}\)
21 tháng 7 2020
b) áp dụng bđt \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)ta có:
\(\hept{\begin{cases}\frac{1}{a+b-c}+\frac{1}{b+c-a}\ge\frac{4}{a+b-c+b+c-a}=\frac{4}{2b}=\frac{2}{b}\\\frac{1}{b+c-a}+\frac{1}{a+c-b}\ge\frac{4}{b+c-a+a+c-b}=\frac{4}{2c}=\frac{2}{c}\\\frac{1}{a+b-c}+\frac{1}{a+c-b}\ge\frac{4}{a+b-c+a+c-b}=\frac{4}{2a}=\frac{2}{a}\end{cases}}\)
Cộng theo vế 3 BĐT ta có:
\(2VT\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2VP\)
\(\Rightarrow VT\ge VP\)
Đẳng thức xảy ra <=> a=b=c
Y
19 tháng 5 2019
Theo BĐT AM-GM :
\(\sqrt{b}=\sqrt{b\cdot1}\le\frac{b+1}{2}\)
\(\Rightarrow\frac{a}{\sqrt{b}}\ge\frac{a}{\frac{b+1}{2}}=\frac{2a}{b+1}\)
Dấu "=" xảy ra \(\Leftrightarrow b=1\)
+ Tương tự ta cm đc :
\(\frac{b}{\sqrt{c}}\ge\frac{2b}{c+1}\). Dấu "=" xảy ra \(\Leftrightarrow c=1\)
\(\frac{c}{\sqrt{a}}\ge\frac{2c}{a+1}\). Dấu "=" xảy ra \(\Leftrightarrow a=1\)
Do đó : \(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\ge2\left(\frac{a}{b+1}+\frac{b}{c+}+\frac{c}{a+1}\right)\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{\left(1+1+1\right)^2}{a+b+b+c+c+a}\)
\(=\frac{3^2}{2\left(a+b+c\right)}=\frac{9}{2.4,5}=\frac{9}{9}=1\)
dấu "=" xảy ra khi \(\frac{1}{a+b}=\frac{1}{b+c}=\frac{1}{c+a}=\frac{1}{3}\)