Question

Tìm giá trị nhỏ nhất với a,b,c >0:

Q=\(\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}+\frac{d}{a+b+c}\)

Answer 1

Áp dụng Bất đẳng thức Bunhia- Copski ta có:

\(Q=\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}+\frac{d}{a+b+c}\)

\(=\left(\frac{a}{b+c+d}+1\right)+\left(\frac{b}{a+c+d}+1\right)+\left(\frac{c}{a+b+d}+1\right)+\left(\frac{d}{a+b+c}+1\right)-4\)

\(=\frac{a+b+c+d}{b+c+d}+\frac{a+b+c+d}{a+c+d}+\frac{a+b+c+d}{a+b+d}+\frac{a+b+c+d}{a+b+c}-4\)

\(=\left(a+b+c+d\right)\left(\frac{1}{b+c+d}+\frac{1}{a+c+d}+\frac{1}{a+b+d}+\frac{1}{a+b+c}\right)-4\ge\left(a+b+c+d\right).\frac{\left(1+1+1+1\right)^2}{\left(b+c+d\right)+\left(a+c+d\right)+\left(a+b+d\right)+\left(a+b+c\right)}-4\)

\(=\left(a+b+c+d\right)\frac{16}{3\left(a+b+c+d\right)}-4\)

\(=\frac{16}{3}-4=\frac{4}{3}\)

Dấu '=' xảy ra khi và chỉ khi \(a=b=c=d\)