Cho a,b là số nguyên dương.CMR:\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
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16 tháng 9 2017
Hơi khó :)) mình ms lớp 8
Ta có : \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}=\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}+\frac{c}{b}\)
\(=\frac{a}{b}+\frac{b}{a}+\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{\frac{a}{c}.\frac{c}{a}}+2\sqrt{\frac{b}{c}.\frac{c}{b}}=6\)(AM - GM) (1)
Ta lại có : \(\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)(AM - GM)
\(\Leftrightarrow2\left(a+b+c\right)\ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\ge\frac{3}{\sqrt[3]{\left(b+c\right)\left(a+c\right)\left(a+b\right)}}\)
\(\Rightarrow2\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)\ge9\)
\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}\ge\frac{9}{2}\)
\(\Leftrightarrow3+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{9}{2}\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{3}{2}\)(2)
Từ (1);(2) \(\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}\ge6+\frac{3}{2}=\frac{15}{2}\)(đpcm)
LV
16 tháng 9 2017
\(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}=\frac{2\left(a+b+c\right)}{\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\ge\frac{3}{\left(\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\right)}\)sai
TN
11 tháng 8 2017
Chuẩn hóa \(a+b+c=3\) thì cần c/m
\(\sqrt{\frac{a}{3-a}}+\sqrt{\frac{b}{3-b}}+\sqrt{\frac{c}{3-c}}\ge\frac{3\sqrt{2}}{2}\)
Ta có BĐT phụ \(\sqrt{\frac{a}{3-a}}\ge\frac{3\sqrt{2}}{8}a+\frac{\sqrt{2}}{8}\)
\(\Leftrightarrow\frac{\frac{3\left(a-1\right)^2\left(3a-1\right)}{32\left(3-a\right)}}{\sqrt{\frac{a}{3-a}}+\frac{3\sqrt{2}}{8}a+\frac{\sqrt{2}}{8}}\ge0\forall0< a< 3\)
Tương tự cho 2 BĐT còn lại cũng có:
\(\sqrt{\frac{b}{3-b}}\ge\frac{3\sqrt{2}}{8}b+\frac{\sqrt{2}}{8};\sqrt{\frac{c}{3-c}}\ge\frac{3\sqrt{2}}{8}c+\frac{\sqrt{2}}{8}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\frac{3\sqrt{2}}{8}\left(a+b+c\right)+\frac{\sqrt{2}}{8}\cdot3=\frac{3\sqrt{2}}{2}\)
TN
20 tháng 2 2018
a) Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ta có:
\(\frac{1}{p-a}+\frac{1}{p-b}\ge\frac{4}{2p-a-b}=\frac{4}{a+b+c-a-b}=\frac{4}{c}\left(p=\frac{a+b+c}{2}\right)\)
Tương tự rồi cộng theo vế:
\(2VT\ge\frac{4}{a}+\frac{4}{b}+\frac{4}{c}=2VP\Leftrightarrow VT\ge VP\)
Dấu "=" khi \(a=b=c\)
b)sai đề
29 tháng 3 2019
a) \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)( luôn đúng )
Dấu "=" xảy ra \(\Leftrightarrow a=b\)
9 tháng 8 2017
Ta có :
\(A+3=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+3\)
\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
\(=2017.\frac{1}{2017}=1\)
\(\Rightarrow A=1-3=-2\)
Theo bđt cô si ta có : \(\hept{\begin{cases}a+b\ge2\sqrt{ab}\\\frac{1}{a}+\frac{1}{b}\ge2\sqrt{\frac{1}{ab}}\end{cases}}\)
=> \(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\cdot\sqrt{ab}\cdot\sqrt{\frac{1}{ab}}=4\sqrt{\frac{ab}{ab}}=4\)
=> \(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\)
=> \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ( Đpcm)
Ta có : \(\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b}\)
\(=\frac{b+a}{ab}-\frac{4}{a+b}\)
\(=\frac{\left(a+b\right)^2-4ab}{ab\left(a+b\right)}\)
\(=\frac{\left(a-b\right)^2}{ab\left(a+b\right)}\ge0\) ( luôn đúng ) ( do a ; b là các số nguyên dương )
\(\Rightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
Dấu " = " xảy ra khi :
\(\hept{\begin{cases}a-b=0\\a;b>0\end{cases}}\Leftrightarrow a=b>0\)
Vậy ....