Cho $x,y,z$ là các số thực dương thỏa mãn $x+y+z=1$. Chứng minh rằng $$\dfrac{x^2+3xy}{x+y}+ \dfrac{y^2+3yz}{y+z}+ \dfrac{z^2+3zx}{z+x} \le 2$$
Chứng minh $\sum \frac{x^2+3xy}{x+y} \le 2$ với $x+y+z=1$
Started By Zaraki, 18-07-2013 - 07:21
#1
Posted 18-07-2013 - 07:21
Zaraki
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#2
Posted 18-07-2013 - 08:01
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Hạ sĩ
ta có $\frac{x^{2}+3xy}{x+y}-x= \frac{2xy}{x+y}\leq 2xy(\frac{1}{4x}+\frac{1}{4y})=\frac{x+y}{2}$
tương tự rồi cộng vào ta đc dpcm
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