Question

\(B=\frac{a+bc}{b+c}+\frac{b+ac}{a+c}+\frac{c+ab}{a+b}\)Cho a,b,c>0 và a+b+c=1 Tìm min 

Answer 1

Ta có \(a+bc=a\left(a+b+c\right)+bc=\left(a+b\right)\left(a+c\right)\)

        \(b+ac=\left(b+a\right)\left(b+c\right)\)

        \(c+ab=\left(a+b\right)\left(c+b\right)\)

Đặt \(a+b=x;b+c=y;a+c=z\)=> \(x+y+z=2\)

Khi đó \(P=\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\)

Áp dụng BĐT cosi \(\frac{xy}{z}+\frac{yz}{x}\ge2y\); \(\frac{yz}{x}+\frac{xz}{y}\ge2z\);\(\frac{xy}{z}+\frac{xz}{y}\ge2z\)

Cộng 3 BĐT trên

=> \(P\ge x+y+z=2\)

Vậy MinP=2 khi a=b=c=1/3