Đáp án:
Giải thích các bước giải:
$\dfrac{a}{b+c}+$ $\dfrac{b}{c+a}+$ $\dfrac{c}{a+b}$
$=\dfrac{a}{b+c}+$ 1$+\dfrac{b}{c+a}+1$ $+\dfrac{c}{a+b}+1-3$
$=\dfrac{a+b+c}{b+c}+$ $\dfrac{a+b+c}{c+a}+$ $\dfrac{a+b+c}{a+b}-3$
$=(a+b+c)(_{}$ $\dfrac{1}{a+b}+$ $\dfrac{1}{b+c}+$ $\dfrac{1}{c+a})-3$
áp dụng bất đẳng thức cô-si cho 3 số : $\dfrac{1}{a}+$ $\dfrac{1}{b}+$ $\dfrac{1}{c}$ $\geq$ $\dfrac{9}{a+b+c}$ ta có :
$\dfrac{1}{a+b}+$ $\dfrac{1}{b+c}+$ $\dfrac{1}{c+a}$ $\geq$ $\dfrac{9}{2(a+b+c)}$
⇒$(a+b+c)(_{}$ $\dfrac{1}{a+b}+$ $\dfrac{1}{b+c}+$ $\dfrac{1}{c+a})-3$ $\geq$ $(a+b+c)._{}$ $\dfrac{9}{2(a+b+c)}-3=$ $\dfrac{9}{2}-3=$ $\dfrac{3}{2}$
Dấu = xảy ra khi a=b=c
Vậy $\dfrac{a}{b+c}+$ $\dfrac{b}{c+a}+$ $\dfrac{c}{a+b}$ $\geq$ $\dfrac{3}{2}$
