2 replies to this topic

[pinocchio1923]

Cho: a, b> 0 và a+b\leq 1 S=\frac{1}{a^{3}+b^{3}}+\frac{1}{a^{2}b}+\frac{1}{ab^{2}}

[VuQuyDat]

 $a+b\leq 1 ;S=\frac{1}{a^{3}+b^{3}}+\frac{1}{a^{2}b}+\frac{1}{ab^{2}}$

[VuQuyDat]

$S=\frac{1}{a^{3}+b^{3}}+\frac{1}{a^{2}b}+\frac{1}{ab^{2}}\geq \frac{1}{(a^2+b^2-ab)(a+b)}+\frac{4}{ab(a+b)}$

$\geq \frac{1}{a^2+b^2-ab}+\frac{1}{ab}+\frac{1}{ab}+\frac{1}{ab}+\frac{1}{ab}\geq \frac{16}{(a+b)^2}+\frac{1}{\frac{(a+b)^2}{4}}\geq 16+4=20$

dau = khi a=b=1/2