

cho ba số thực a,b,c >0 thỏa mãn a+b+c=2019. Chứng minh:
[tex]\frac{a}{a+\sqrt{2019a+bc}} +\frac{b}{b+\sqrt{2019b+ca}}+ \frac{c}{c+\sqrt{2019c+ab}} \leq 1[/tex]
[tex]\frac{a}{a+\sqrt{2019a+bc}} +\frac{b}{b+\sqrt{2019b+ca}}+ \frac{c}{c+\sqrt{2019c+ab}} \leq 1[/tex]