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[hungpronc1]

cho x,y,z dương tm : x(x-1)+y(y-1)+z(z-1)$\leqslant 6$

Tìm Min: P=$\frac{1}{x+y+1}$+$\frac{1}{z+y+1}+\frac{1}{z+x+1}$

[nguyenthehoan]

Theo bài ra 

 

$\sum x^{2}\leq \sum x+6\Rightarrow \sum x+6\geq \frac{(\sum x)^{2}}{3}$

 

$\Rightarrow x+y+z\leq 6$

 

Khi đó 

 

$P=\sum \frac{1}{a+b+1}\geq \frac{9}{2(a+b+c)+3}\geq \frac{3}{5}$

 

Vậy minP=$\frac{3}{5}$