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Bài này có bạn giải rồi:
Cho các số thực dương a,b,c.Chứng minh rằng :\(\dfrac{b\left(2a-b\right)}{a\left(b+c\right)}+\dfrac{c\left(2b-c\right)}{... - Hoc24
\(\dfrac{bc}{a+b+c+a}\le\dfrac{bc}{4}\cdot\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\ \dfrac{ac}{b+c+a+b}\le\dfrac{ac}{4}\cdot\left(\dfrac{1}{b+c}+\dfrac{1}{a+b}\right)\\ \dfrac{ab}{a+c+b+c}\le\dfrac{ab}{4}\cdot\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\\ \Leftrightarrow VT\le\dfrac{1}{a+b}\left(\dfrac{bc}{4}+\dfrac{ac}{4}\right)+\dfrac{1}{a+c}\left(\dfrac{bc}{4}+\dfrac{ab}{4}\right)+\dfrac{1}{b+c}\left(\dfrac{ac}{4}+\dfrac{ab}{4}\right)\\ =\dfrac{1}{a+b}\cdot\dfrac{c\left(a+b\right)}{4}+\dfrac{1}{a+c}\cdot\dfrac{b\left(a+c\right)}{4}+\dfrac{1}{b+c}\cdot\dfrac{a\left(b+c\right)}{4}\\ =\dfrac{c}{4}+\dfrac{b}{4}+\dfrac{a}{4}\\ =\dfrac{a+b+c}{4}\left(đfcm\right)\)
a.
\(\sum\dfrac{ab}{a+c+b+c}\le\dfrac{1}{4}\sum\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)=\dfrac{a+b+c}{4}\)
2.
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{a+b+2c+2b}\le\dfrac{ab}{9}\left(\dfrac{4}{a+b+2c}+\dfrac{1}{2b}\right)=4.\dfrac{ab}{a+b+2c}+\dfrac{a}{18}\)
Quay lại câu a
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2a+b+c}=\dfrac{a}{a+b+c+a}\le\dfrac{a}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{c+a}\right)\\\dfrac{b}{a+2b+c}=\dfrac{b}{a+b+b+c}\le\dfrac{b}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\\\dfrac{c}{a+b+2c}=\dfrac{c}{a+c+b+c}\le\dfrac{c}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{a}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{b}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)+\dfrac{c}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\)
\(\Rightarrow VT\le\dfrac{a}{4\left(a+b\right)}+\dfrac{a}{4\left(a+c\right)}+\dfrac{b}{4\left(a+b\right)}+\dfrac{b}{4\left(b+c\right)}+\dfrac{c}{4\left(a+c\right)}+\dfrac{c}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\left[\dfrac{a}{4\left(a+b\right)}+\dfrac{b}{4\left(a+b\right)}\right]+\left[\dfrac{b}{4\left(b+c\right)}+\dfrac{c}{4\left(b+c\right)}\right]+\left[\dfrac{c}{4\left(a+c\right)}+\dfrac{a}{4\left(a+c\right)}\right]\)
\(\Rightarrow VT\le\dfrac{a+b}{4\left(a+b\right)}+\dfrac{b+c}{4\left(b+c\right)}+\dfrac{c+a}{4\left(c+a\right)}\)
\(\Rightarrow VT\le\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{a}{2a+b+c}+\dfrac{b}{a+2b+c}+\dfrac{c}{a+b+2c}\le\dfrac{3}{4}\) ( đpcm )
Dấu "=" xảy ra khi \(a=b=c\)
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