Question

[Quốc Bảo]

Cho a , b , c > 0 Chứng minh rằng: \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\le\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Answer 1

[Kuro Kazuya]

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 )

Answer 2

[katherina]

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)