khanhlinh8b

$S=\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{c^2}}+\sqrt{c^2+\frac{1}{a^2}}\\\ge\sqrt{(a+b+c)^2+(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}\\\ge\sqrt{(a+b+c)^2+\frac{81}{(a+b+c)^2}}\\=\sqrt{(a+b+c)^2+\frac{81}{16(a++c)^2}+\frac{1215}{16(a+b+c)^2}}\\\ge\sqrt{2\sqrt{\frac{81}{16}}+\frac{1215}{16.(\frac{3}{2})^2}}=\frac{3\sqrt{17}}{2}$

 Cái này là dùng BĐT gì vậy bạn?

$M=\sum \frac{ab}{\sqrt{ab+2c}}=\sum \frac{ab}{\sqrt{ab+c(a+b+c)}}=\sum \frac{ab}{\sqrt{(a+c)(b+c)}}\leq \sum \frac{ab}{2}\left ( \frac{1}{a+c}+\frac{1}{b+c} \right )=\sum \frac{1}{2}\left ( \frac{ab}{a+c}+\frac{ab}{b+c} \right )= \frac{1}{2}\left ( \sum \frac{a(b+c)}{b+c} \right )=1$

Bạn giải thích thêm cho mình dòng này được k?