Question

Cho a,b > 0

Chứng ming rằng: \(\dfrac{1}{a^3}\)+\(\dfrac{a^3}{b^3}\)+b^{3\(\ge\dfrac{1}{a}\)}+\(\dfrac{a}{b}\)+b

Answer 1

Lời giải:

Áp dụng BĐT Cauchy cho $3$ số:

\(\left\{\begin{matrix} \frac{1}{a^3}+1+1\geq \frac{3}{a}\\ \frac{a^3}{b^3}+1+1\geq \frac{3a}{b}\\ b^3+1+1\geq 3b\end{matrix}\right.\Rightarrow \text{VT}\geq 3\text{VP}-6\)

Cũng áp dụng Cauchy:

\(\frac{1}{a}+\frac{a}{b}+b\geq 3\sqrt[3]{\frac{ab}{ab}}=3\Leftrightarrow \text{VP}\geq 3\)

\(\Rightarrow \text{VT}\geq 3\text{VP}-6\geq \text{VP}\) (đpcm)

Dấu bằng xảy ra khi \(a=b=1\)