Question

Answer 1

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:
\(S=\frac{1}{a^3+b^3}+\frac{1}{2a^2b}+\frac{1}{2a^2b}+\frac{1}{2ab^2}+\frac{1}{2ab^2}\)

\(\geq \frac{25}{a^3+b^3+4a^2b+4ab^2}=\frac{25}{(a+b)^2+ab(a+b)}\geq \frac{25}{1+ab}\) do $a+b\leq 1$

Mà áp dụng BĐT Cô-si: $1\geq a+b\geq 2\sqrt{ab}$

$\Rightarrow ab\leq \frac{1}{4}$

$\Rightarrow S\geq \frac{25}{1+\frac{1}{4}}=20$

Vậy $S_{\min}=20$ khi $a=b=\frac{1}{2}$