Question

cho a,b,c>0 thỏa mã a+b+c=6. Tìm Min A=\(\dfrac{2a}{b^2+2}+\dfrac{2b}{c^2+2}+\dfrac{2c}{a^2+2}\)

Answer 1

Lời giải:

Áp dụng BĐT AM-GM:
\(A=\sum \frac{2a}{b^2+2}=\sum (a-\frac{ab^2}{b^2+2})=\sum a-\sum \frac{ab^2}{b^2+2}\)

\(=6-\sum \frac{ab^2}{b^2+2}=6-\sum \frac{ab^2}{\frac{b^2}{2}+\frac{b^2}{2}+2}\)

\(\geq 6-\sum \frac{ab^2}{3\sqrt[3]{\frac{b^4}{2}}}=6-\frac{1}{3}\sum \sqrt[3]{2a^3b^2}\)

Tiếp tục áp dụng BĐT AM-GM:

\(\sum \sqrt[3]{2a^3b^2}\leq \sum \frac{2a+ab+ab}{3}=\frac{12+2(ab+bc+ac)}{3}=4+\frac{2}{3}(ab+bc+ac)\)

\(\leq 4+\frac{2}{3}.\frac{(a+b+c)^2}{3}=12\)

Do đó: $A\geq 6-\frac{1}{3}.12=2$

Vậy $A_{\min}=2$ khi $a=b=c=2$