What is the smallest real number \(k\) (to 3 decimal places), such that for all ordered triples of non-negative reals \( (a,b,c) \) which satisfy \( a + b + c = 1 \), we have

\[ \frac{ a^2 + bc} { a+1} + \frac{ b^2+ca} { b+1} + \frac{ c^2+ab} { c+1} \leq k? \]

If you want a similar inequality, you can try The Answer Is Not 0.225.

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