Question

Cho 2 số thực dương a, b thỏa mãn a+b\(\le\)1

a) B=\(\frac{1}{a^2+b^2}+\frac{1}{ab}+4ab\)

b) C=\(\frac{1}{a^3+b^3}+\frac{1}{a^2b}+\frac{1}{ab^2}\)

Answer 1

a)\(B=\frac{1}{a^2+b^2}+\frac{1}{ab}+4ab=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{2ab}+8ab-4ab\)

Áp dụng BĐT AM-GM ta có:

\(B=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{2ab}+8ab-4ab\)

\(\ge\frac{4}{\left(a+b\right)^2}+2\sqrt{\frac{1}{2ab}\cdot8ab}-\left(a+b\right)^2=7\)

Dấu "=" xảy ra khi \(\begin{cases}a=b\\a+b=1\end{cases}\)\(\Rightarrow a=b=\frac{1}{2}\)

Vậy \(Min_B=7\) khi \(a=b=\frac{1}{2}\)

b)\(C\ge\frac{1}{1-3ab\left(a+b\right)}+\frac{4}{ab\left(a+b\right)}\)

\(\ge\frac{16}{1-3ab\left(a+b\right)+3ab\left(a+b\right)}+\frac{1}{\frac{\left(a+b\right)^3}{4}}\ge16+4=20\)

Dấu "=" xảy ra khi \(\begin{cases}a=b\\a+b=1\end{cases}\)\(\Rightarrow a=b=\frac{1}{2}\)

Vậy \(Min_C=20\) khi \(a=b=\frac{1}{2}\)