<s>[INST] Whenever $a+b+c=1$, $\frac{bc+a+1}{a^2+1} + \frac{ac+b+1}{b^2+1} + \frac{ab+c+1}{c^2+1} \le \frac{39}{10}$ Prove that for a + b + c =1 and a,b,c are positive real numbers, 
then
$$\frac{bc+a+1}{a^2+1} + \frac{ac+b+1}{b^2+1} + \frac{ab+c+1}{c^2+1} \le \frac{39}{10}$$
My try: 
if one term is proven to be $\le \frac{13}{10} $ then we prove the inequality.
sub in $a =1-b-c$
then we have
$$\frac{bc+b+c+2}{b^2+c^2+2bc-2b-2c+2}\le\frac{13}{10}$$
further simplifying, then we have $$0\le13b^2+13c^2+16bc-16b-16c+6$$
since  $b^2+c^2\ge 2bc$ then we can substitute it? (not sure about this)
then we have
$$0\le42bc-16b-16c+6$$ since $b+c\ge2\sqrt{bc}$ then we can sub it again? (also not sure about this)
then we get a quadratic inequality
$$21bc -16\sqrt{bc} +6$$
but it is not true for all positive real numbers a, b and c.
Is it impossible to use this approach and get the answer (is there a better way to solve it) or did i do something wrong? 
 [/INST] This is wrong. If $a=1$ and $b=c=0$ then the left-hand side is $3$, which is less than $\frac{39}{10}$.
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