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Show that for any positive $a,b,c$ and any $\alpha \geq 65/66$ we have $$\alpha\left( \frac{1}{a b} + \frac{1}{(b+c)(a+b+c)} + \frac{1}{(a+b+c)(a+b)} + \frac{1}{b c}\right)+ \frac{1}{(b+c)^2} + \frac{1}{(a+b+c)^2}$$ $$ +\frac{1}{(a+b)^2} + \left( 1 - 2\alpha \right) \left( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right) \leq 0.$$

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The inequality is false, e.g., for $a=63$, $b=65$, $c=84$, $\alpha=14074/14269$.

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