Question

Answer 1

Lời giải:

$a+b+c=abc$

$\Rightarrow a(a+b+c)=a^2bc$

$\Rightarrow a(a+b+c)+bc=bc(a^2+1)$
$\Rightarrow bc(a^2+1)=(a+b)(a+c)\Rightarrow a^2+1=\frac{(a+b)(a+c)}{bc}$

Hoàn toàn tương tự: $b^2+1=\frac{(b+a)(b+c)}{ac}$

$c^2+1=\frac{(c+a)(c+b)}{ab}$

Do đó:

\(\text{VT}=\sum \frac{1}{\sqrt{\frac{(a+b)(a+c)}{bc}}}=\sum \sqrt{\frac{bc}{(a+b)(a+c)}}\leq \sum \frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\) (theo BĐT AM-GM)

\(\Rightarrow \text{VT}\leq \frac{1}{2}(\frac{b+a}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a})=\frac{3}{2}\)

Vậy ta có đpcm

Dấu "=" xảy ra khi $a=b=c=\sqrt{3}$