Chủ đề này có 2 trả lời

[ngonluahoangkim]

cho các số thực dương a,b,c thoả mãn abc=1. chứng minh

             $\frac{b+c}{\sqrt{a}}$+$\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$

 

[Khoa Linh]

$\sum \left (\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{a}} \right )\geq \frac{\left ( 2\left ( \sqrt{a}+\sqrt{b}+\sqrt{c} \right ) \right )^2}{2\sqrt{a}+2\sqrt{b}+2\sqrt{c}}=\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b}+\sqrt{c}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$

[DOTOANNANG]

Từ BĐT AM- GM, ta có:

$\frac{b+ c}{\sqrt{a}}+ \frac{c+ a}{\sqrt{b}}+ \frac{a+ b}{\sqrt{c}}\geq 2\left ( \sqrt{\frac{bc}{a}}+ \sqrt{\frac{ca}{b}}+ \sqrt{\frac{ab}{c}} \right )= \left ( \sqrt{\frac{bc}{a}}+ \sqrt{\frac{ca}{b}} \right )+ \left ( \sqrt{\frac{ca}{b}}+ \sqrt{\frac{ab}{c}} \right )+ \left ( \sqrt{\frac{ab}{c}}+ \sqrt{\frac{bc}{a}} \right )\geq 2\left ( \sqrt{a}+ \sqrt{b}+ \sqrt{c} \right )\geq \sqrt{a}+ \sqrt{b}+ \sqrt{c}+ 3\sqrt[6]{abc}= \sqrt{a}+ \sqrt{b}+ \sqrt{c}+ 3$