đề có sai 1 chút nha bạn :
đề phải là \(a;b;c>0\) : \(CMR\) \(\dfrac{a}{b+c}+\dfrac{9b}{a+c}+\dfrac{16c}{a+b}\ge6\) mới đúng
giải
đặt : \(P=\dfrac{a}{b+c}+\dfrac{9b}{a+c}+\dfrac{16c}{a+b}\)
ta có : \(P=\dfrac{a}{b+c}+\dfrac{9b}{a+c}+\dfrac{16c}{a+b}\)
\(P=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{9b}{a+c}+9\right)+\left(\dfrac{16c}{a+b}+16\right)-26\)
\(P=\left(\dfrac{a+b+c}{b+c}\right)+\left(\dfrac{9b+9a+9c}{a+c}\right)+\left(\dfrac{16c+16a+16b}{a+b}\right)-26\)
\(P=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{9}{a+c}+\dfrac{16}{a+b}\right)-26\)
\(P=\dfrac{1}{2}\left(\left(b+c\right)+\left(a+c\right)+\left(a+b\right)\right)\left(\dfrac{1}{b+c}+\dfrac{9}{a+c}+\dfrac{16}{a+b}\right)-26\)
áp dụng bất đẳng thức Bunhiacopxki
ta có :
\(\left(\left(b+c\right)+\left(a+c\right)+\left(a+b\right)\right)\left(\dfrac{1}{b+c}+\dfrac{9}{a+c}+\dfrac{16}{a+b}\right)\ge\left(\sqrt{1}+\sqrt{9}+\sqrt{16}\right)^2\)
\(\Leftrightarrow\left(\left(b+c\right)+\left(a+c\right)+\left(a+b\right)\right)\left(\dfrac{1}{b+c}+\dfrac{9}{a+c}+\dfrac{16}{a+b}\right)\ge64\)
\(\Leftrightarrow\) \(P=\dfrac{1}{2}\left(\left(b+c\right)+\left(a+c\right)+\left(a+b\right)\right)\left(\dfrac{1}{b+c}+\dfrac{9}{a+c}+\dfrac{16}{a+b}\right)-26\ge\dfrac{1}{2}.64-26\)
\(\Leftrightarrow P\ge6\)
vậy \(P=\dfrac{a}{b+c}+\dfrac{9b}{a+c}+\dfrac{16c}{a+b}\ge6\) (đpcm)
dấu "=" xảy ra khi \(b+c=\dfrac{a+c}{9}=\dfrac{a+b}{16}\)
