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What is the smallest real number kkk (to 3 decimal places), such that for all ordered triples of non-negative reals (a,b,c) (a,b,c) (a,b,c) which satisfy a+b+c=1 a + b + c = 1 a+b+c=1, we have
a2+bca+1+b2+cab+1+c2+abc+1≤k? \frac{ a^2 + bc} { a+1} + \frac{ b^2+ca} { b+1} + \frac{ c^2+ab} { c+1} \leq k? a+1a2+bc+b+1b2+ca+c+1c2+ab≤k?
If you want a similar inequality, you can try The Answer Is Not 0.225.
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