Ta có $a+b+c=3\geq ab+bc+ca$. NênĐưa tiếp 1 bài lên vậy
Bài 10: Cho $a,b,c>0;a+b+c=3$.Chứng minh rằng:
$\dfrac{bc}{\sqrt{a^2+3}}+\dfrac{ac}{\sqrt{b^2+3}}+\dfrac{ab}{\sqrt{c^2+3}} \le \dfrac{3}{2}$
$\dfrac{bc}{\sqrt{a^2+3}}+\dfrac{ac}{\sqrt{b^2+3}}+\dfrac{ab}{\sqrt{c^2+3}} \le \sum \dfrac{bc}{\sqrt{(a+c)(a+b)}}\leq \sum \dfrac{1}{2}\left ( \dfrac{bc}{a+b}+\dfrac{ac}{a+b} \right )=\dfrac{3}{2}$