Question

Cho ba số dương a,b,c . Chứng minh rằng
\(\dfrac{a}{b+c}+\dfrac{4b}{c+a}+\dfrac{9c}{a+b}>4\)
 

Answer 1

\(VT=\dfrac{a}{b+c}+1+\dfrac{4b}{c+a}+4+\dfrac{9c}{a+b}+9-14\)

\(VT=\dfrac{a+b+c}{b+c}+\dfrac{4\left(a+b+c\right)}{c+a}+\dfrac{9\left(a+b+c\right)}{a+b}-14\)

\(VT=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{4}{c+a}+\dfrac{9}{a+b}\right)-14\)

\(VT\ge\left(a+b+c\right).\dfrac{\left(1+2+3\right)^2}{b+c+c+a+a+b}-14=4\) 

Dấu "=" không xảy ra nên \(VT>4\) (đpcm)