\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{4c}{a+b}\ge2\)
\(VT:\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{4c}{a+b}\)
\(=\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1+\dfrac{4c}{a+b}+4-6\\ =\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{4a+4b+4c}{a+b}-6\\ =\left(a+b+c\right)\cdot\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{4}{a+b}\right)-6\)
Áp dụng bđt Caychuy - Schwarz :
\ \(\left(a+b+c\right)\cdot\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{4}{a+b}\right)-6\\ \ge\left(a+b+c\right)\cdot\dfrac{\left(1+1+2\right)^2}{2a+2b+2c}-6\\ \ge\dfrac{16}{2}-6=2\)
Dấu = xảy ra khi \(a=b=c\)
Cách khác:
Ta có: \(\left(a-b\right)^2+4c^2\ge0\Leftrightarrow a^2-2ab+b^2+4c^2\ge0\)
\(\Leftrightarrow a^2+b^2+4c^2+2ab+4bc+4ac-4ab-4bc-4ca\ge0\)\(\Leftrightarrow\left(a+b+2c\right)^2-4\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow\left(a+b+2c\right)^2\ge4\left(ab+bc+ca\right)\)
Theo BĐT cauchy-schwarz ta có:
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{4c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{4c^2}{ac+ab}\ge\)\(\ge\dfrac{\left(a+b+2c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{4\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=2\Rightarrowđpcm\)
