Cyclic inequality for three positive numbers

For positive $a$, $b$, $c$, I would like to know if the following inequality is true $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} \leq 3$$ I appreciate all comments on this.

• No: $a=10$, $b=c=1$. – njguliyev Aug 30 '13 at 13:16
• More often than not, these inequalities come with a second condition, like $abc = 1$ or $a + b + c \leq 1$. Are you certain you didn't miss something like that? – Arthur Aug 30 '13 at 13:18
• Fix $a$ and $b$ and increase $c$ as much as you like. The third term is unbounded, and the others remain positive. – Mark Bennet Aug 30 '13 at 13:25
• Maybe it's false.Try $a\rightarrow \infty,b,c\rightarrow 0$ – Smy2012 Aug 30 '13 at 14:59
• Thank you for your replies. Maybe with the additional condition that $a, b, c$ are sides of a triangle the inequality is true. I see that the given counterexamples don't cover that case. – Martin Aug 31 '13 at 5:05

Suppose $b=a$ and $c=3a$. The inequality reduces to the following:
$\frac{a}{4a}+\frac{a}{4a}+\frac{3a}{2a}+\frac{a^2}{2a^2}+\frac{3a^2}{10a^2}+\frac{3a^2}{10a^2} \leq 3$
$\frac{1}{4}+\frac{1}{4}+\frac{3}{2}+\frac{1}{2}+\frac{3}{10}+\frac{3}{10} \leq 3$
$\frac{31}{10} \leq 3$
$3.1 \leq 3$
Contradiction, so there must be infinitely many tuples $\left( a, b, c \right)$ such that the inequality does not hold.