Giả sử điều cần chứng minh đúng thì:
\(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)
\(\Rightarrow\dfrac{a}{b+c}+1+\dfrac{b}{a+c}+1+\dfrac{c}{a+b}+1\ge\dfrac{3}{2}+1+1+1=\dfrac{9}{2}\)
\(\Rightarrow\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}\ge\dfrac{9}{2}\)
\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\ge\dfrac{9}{2}\)
\(\Rightarrow\Rightarrow2\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\ge9\)
\(\Rightarrow\left(a+b+c+a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\ge9\)
Đặt: \(\left\{{}\begin{matrix}b+c=x\\a+c=y\\a+b=z\end{matrix}\right.\) Khi đó:
\(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\)(đúng theo AM-GM)
Ta có đpcm
Rối'ss :v
Đặt VT là A\(A=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel, ta có:
\(A\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)
Lưu ý: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)(Dễ dàng chứng minh bđt này nhờ Cauchy hoặc hằng đẳng thức)
