Ta co BĐT sau đây: $\sqrt{1+x^3}\leq 1+\frac{x^2}{2}$Thật vậy: $\sqrt{1+x^3}=\sqrt{(1+x)(1-x+x^2)}\leq \frac{1+x+1-x+x^2}{2}=1+\frac{x^2}{2}$
Vậy $S\geq \frac{a}{1+\frac{b^2}{2}}+\frac{b}{1+\frac{a^2}{2}}=\frac{a^2}{a+\frac{ab^2}{2}}+\frac{b^2}{b+\frac{a^2b}{2}}$
áp dụng cau chy schwarz$\Rightarrow S\geq \frac{(a+b)^2}{a+b+ab(\frac{a+b}{2})}\geq \frac{(a+b)^2}{a+b+(\frac{a+b}{2})^3}=\frac{4}{3}$
Vậy $Min S=\frac{4}{3}$ xảy ra tại $a=b=2$