Question

Cho a,b,c là ba số thực dương. Chứng minh rằng \(\frac{5a+c}{b+c}+\frac{6b}{c+a}+\frac{5c+a}{a+b}\ge9\)

Answer 1

\(VT=\frac{\left(5a+c\right)^2}{\left(b+c\right)\left(5a+c\right)}+\frac{\left(6b\right)^2}{6b\left(a+c\right)}+\frac{\left(5c+a\right)^2}{\left(a+b\right)\left(5c+a\right)}\)

\(VT\ge\frac{\left(5a+c+6b+5c+a\right)^2}{5ab+5ac+bc+c^2+6ab+6bc+5ac+5bc+a^2+ab}\)

\(VT\ge\frac{36\left(a+b+c\right)^2}{a^2+c^2+12ab+12bc+10ac}\ge\frac{36\left(a+b+c\right)^2}{a^2+c^2+a^2+b^2+b^2+c^2+10ab+10bc+10ac}\)

\(VT\ge\frac{36\left(a+b+c\right)^2}{2\left(a+b+c\right)^2+6\left(ab+bc+ca\right)}\ge\frac{36\left(a+b+c\right)^2}{2\left(a+b+c\right)^2+2\left(a+b+c\right)^2}=9\)

Dấu "=" xảy ra khi \(a=b=c\)