Cho x,y,z dương thỏa mãn: $\frac{3x}{x+1}+\frac{4y}{y+1}+\frac{2z}{z+1}=2$
Chứng minh $x^3y^4z^2\leqslant \frac{1}{8^9}$
Chứng minh $x^3y^4z^2\leqslant \frac{1}{8^9}$
Started By alexander123, 15-12-2021 - 18:06
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