Question

Answer 1

Lời giải:

Áp dụng BĐT AM-GM:

$3\geq (a+b+c)^2\geq 3(ab+bc+ac)\Rightarrow ab+bc+ac\leq 1$

Khi đó:
$a^2+1\geq a^2+ab+bc+ac=(a+b)(a+c)$

$b^2+1\geq b^2+ab+bc+ac=(b+a)(b+c)$

$c^2+1\geq c^2+ab+bc+ac=(c+a)(c+b)$

Tiếp tục áp dụng BĐT AM-GM:

\(\sum \frac{a}{\sqrt{1+a^2}}\leq \sum \frac{a}{\sqrt{(a+b)(a+c)}}\leq \sum \frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{1}{2}(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a})=\frac{3}{2}\)

Ta có đpcm.

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