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Posted by gicungduoc on 03-06-2018 - 22:30
cho ba số thực a,b,c thỏa mãn a+b+c=2019.chứng minh:
$\frac{a}{{a+\sqrt{2019a+bc}}}+\frac{b}{b+\sqrt{2019b+ca}}+\frac{c}{c+\sqrt{2019c+ab}}\leq 1$