Question

cho các số dương a,b,c.chứng minh bất đẳng thức:
\(\dfrac{a+b}{c}\)+\(\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge4(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b})\)

Answer 1

\(VP=\dfrac{4a}{b+c}+\dfrac{4b}{c+a}+\dfrac{4c}{a+b}\)

Áp dụng BĐT \(\dfrac{4}{x+y}\le\dfrac{1}{x}+\dfrac{1}{y}\Leftrightarrow\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\) có:

\(\dfrac{4a}{b+c}\le\dfrac{1}{4}\left(\dfrac{4a}{b}+\dfrac{4a}{c}\right)=\dfrac{4a}{b}\cdot\dfrac{1}{4}+\dfrac{4a}{c}\cdot\dfrac{1}{4}=\dfrac{a}{b}+\dfrac{a}{c}\)

Tương tự cho 2 BĐT còn lại ta có:

\(\dfrac{4b}{a+c}\le\dfrac{b}{a}+\dfrac{b}{c};\dfrac{4c}{a+b}\le\dfrac{c}{a}+\dfrac{c}{b}\)

Cộng theo vế 3 BĐT trên ta có:

\(\dfrac{4a}{b+c}+\dfrac{4b}{c+a}+\dfrac{4c}{a+b}\le\left(\dfrac{a}{c}+\dfrac{b}{c}\right)+\left(\dfrac{b}{a}+\dfrac{c}{a}\right)+\left(\dfrac{c}{b}+\dfrac{a}{b}\right)\)​

\(\Leftrightarrow\dfrac{4a}{b+c}+\dfrac{4b}{c+a}+\dfrac{4c}{a+b}\le\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\)

​

Answer 2

Ta có: \(\dfrac{a+b}{c}\)+\(\dfrac{b+c}{a}\)+\(\dfrac{c+a}{b}\)=\(\dfrac{a+b}{c}\)+\(\dfrac{c}{a+b}\)+\(\dfrac{b+c}{a}\)+\(\dfrac{a}{b+c}\)+\(\dfrac{c+a}{b}\)+\(\dfrac{b}{c+a}\)-(\(\dfrac{c}{a+b}\)+\(\dfrac{a}{c+b}\)+\(\dfrac{b}{a+c}\))

a/d bdt cosi cho...........................ta có

A\(\ge\)2\(\sqrt{\dfrac{a+b}{c}\times\dfrac{c}{a+b}}\)+

2\(\sqrt{\dfrac{b+c}{a}\times\dfrac{a}{b+c}}\)+2\(\sqrt{\dfrac{a+c}{b}\times\dfrac{b}{a+c}}\)

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