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Positive real numbers aaa, bbb, and ccc are such that 1a+1b+1c=a+b+c\dfrac{1}{a}+ \dfrac{1}{b}+ \dfrac{1}{c} =a+b+ca1+b1+c1=a+b+c. And if
1(2a+b+c)2+1(a+2b+c)2+1(a+b+2c)2≤pq\frac{1}{(2a+b+c)^2}+ \frac{1}{(a+2b+c)^2}+ \frac{1}{(a+b+2c)^2} \leq \dfrac{p}{q} (2a+b+c)21+(a+2b+c)21+(a+b+2c)21≤qp
where ppp and qqq are coprime positive integers. Find p+qp+qp+q.
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