\(P=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(=\frac{a^2}{ab+ac}+\frac{b^2}{bc+ba}+\frac{c^2}{ac+bc}\)
\(\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
\(\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)
dấu "=" xảy ra tại a=b=c
Cách 2
\(P+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge\left(a+b+c\right)\cdot\frac{9}{2\left(a+b+c\right)}=\frac{9}{2}\)
\(\Rightarrow P\ge\frac{3}{2}\Leftrightarrow a=b=c\)
\(P=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\left(1\right)\)
Đặt \(\hept{\begin{cases}b+c=x\\c+a=y\\a+b=z\end{cases}\left(x,y,z>0\right)}\)
\(\Rightarrow a=\frac{y+z-x}{2}\);\(b=\frac{z+x-y}{2}\);\(c=\frac{x+y-z}{2}\)
\(\left(1\right)\)trở thành \(\frac{y+z-x}{2x}+\frac{z+x-y}{2y}+\frac{x+y-z}{2z}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{y}{2x}+\frac{z}{2x}-\frac{1}{2}+\frac{z}{2y}+\frac{x}{2y}-\frac{1}{2}+\frac{x}{2z}+\frac{y}{2z}-\frac{1}{2}\ge\frac{3}{2}\)
\(\Leftrightarrow\left(\frac{y}{2x}+\frac{x}{2y}\right)+\left(\frac{z}{2x}+\frac{x}{2z}\right)+\left(\frac{z}{2y}+\frac{y}{2z}\right)\ge3\)
Vì \(\frac{y}{2x}+\frac{x}{2y}\ge2\sqrt{\frac{y}{2x}.\frac{x}{2y}}=1\)( bđt AM-GM)
CMTT \(\frac{z}{2x}+\frac{x}{2z}\ge1\)và \(\frac{z}{2y}+\frac{y}{2z}\ge1\)
rồi cộng vào là xong
Dấu"="xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{y}{2x}=\frac{x}{2y}\\\frac{z}{2x}=\frac{x}{2z}\\\frac{z}{2y}=\frac{y}{2z}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}2x^2=2y^2\\2z^2=2x^2\\2y^2=2z^2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=y\\z=x\\y=z\end{cases}\Leftrightarrow}x=y=z\)
Vậy \(P_{min}=\frac{3}{2}\Leftrightarrow x=y=z\)
