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Let a, b, c be positive real numbers such that \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = a+b+c. Prove that:
\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}


Proposed by Juhan Aru, Estonia

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