Câu hỏi của Quốc Bảo

Question

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

Kuro Kazuya · Accepted Answer

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 )Câu trả lời: 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 )

katherina · Answer

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)Câu trả lời: Với a , b , c > 0