Chủ đề này có 3 trả lời

[DOTOANNANG]

\[a, b, c> 0\]

\[a+ b+ c= 1\]

CM: \[\sqrt{1+ \frac{a}{bc}}+ \sqrt{1+ \frac{b}{ca}}+ \sqrt{1+ \frac{c}{ab}}\geq 6\]

 

[Khoa Linh]

\[a, b, c> 0\]

\[a+ b+ c= 1\]

CM: \[\sqrt{1+ \frac{a}{bc}}+ \sqrt{1+ \frac{b}{ca}}+ \sqrt{1+ \frac{c}{ab}}\geq 6\]

 

Ta có: a+b+c=1 nên dễ CM được $abc\leq \frac{1}{27}$

 

 

Áp dụng BĐT minkowski ta có:

$\sqrt{1+ \frac{a}{bc}}+ \sqrt{1+ \frac{b}{ca}}+ \sqrt{1+ \frac{c}{ab}}\geq \sqrt{(1+1+1)^2+\left ( \sqrt{\frac{a}{bc}}+\sqrt{\frac{b}{ca}}+\sqrt{\frac{c}{ab}} \right )^2}\geq \sqrt{9+9.\sqrt[3]{\frac{1}{abc}}}\geq \sqrt{9+9.3}=6$

[nmtuan2001]

\[a, b, c> 0\]

\[a+ b+ c= 1\]

CM: \[\sqrt{1+ \frac{a}{bc}}+ \sqrt{1+ \frac{b}{ca}}+ \sqrt{1+ \frac{c}{ab}}\geq 6\]

Cách khác: Nhân cả 2 vế với $\sqrt{abc}$, BĐT tương đương với

$$\sum \sqrt{a(a+bc)} \geq 6\sqrt{abc}$$

Áp dụng BĐT Cauchy-Schwarz:

$$\sum \sqrt{a(a+bc)}=\sum \sqrt{a[a(a+b+c)+bc]}=\sum \sqrt{a(a+b)(a+c)}$$

$$\geq \sum \sqrt{a}(a+\sqrt{bc})=\sum a\sqrt{a}+3\sqrt{abc} \geq 3\sqrt{abc}$$

[melodias2002]

\[a, b, c> 0\]

\[a+ b+ c= 1\]

CM: \[\sqrt{1+ \frac{a}{bc}}+ \sqrt{1+ \frac{b}{ca}}+ \sqrt{1+ \frac{c}{ab}}\geq 6\]

$\sum \sqrt{1+\frac{a}{bc}} = \sum \sqrt{\frac{a+bc}{bc}} = \sum \sqrt {\frac{(a+b)(a+c)}{bc}} \geq 3\sqrt[3]{\prod \sqrt{\frac{(a+b)(a+c)}{bc}}}=3\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{abc}}\geq3\sqrt[3]{8}=6$

Dấu "=" xảy ra $\Leftrightarrow a=b=c=\frac{1}{3}$