Question

Cho a,b,c>0. Tìm Min P=\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Q=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

Answer 1

\(P=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)\)

Ta có: \(\frac{a}{b}+\frac{b}{a}\ge2;\frac{b}{c}+\frac{c}{b}\ge2;\frac{c}{a}+\frac{a}{c}\ge2\)

\(\Rightarrow P\ge3+2+2+2=9\)

\("="\Leftrightarrow a=b=c\)

\(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\ge\frac{3\left(ab+bc+ac\right)}{2\left(ab+bc+ac\right)}=\frac{3}{2}\)

\("="\Leftrightarrow a=b=c\)