Inequality. $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2} \geq 3+n$ I can't find any solution for this inequality which can be found here
Exercise 1.3.5 Let $a,b,c$ be positive real numbers such that $a+b+c=ab+bc+ca$ and $n \leq 3$. Prove that 

$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2} \geq 3+n$$

Thanks :)

Making use of given hint, 
$$ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq {(a + b +c)(a^2 + b^2 + c^2)\over ab + bc +ac} = {(a + b +c)(a^2 + b^2 + c^2)\over (a+b+c)} = a^2 + b^2 +c^2$$
The minimum value of $a^2+b^2+c^2 \geq a+b+c \geq 3 $ for given criteria. 
Let ${a^2 + b^2 + c^2 = z}$ then $f(z, n) = z + {3n \over z}$ for $n \leq 3$ and $z \geq 3$
$$f(z,n) = z + {3n \over z} \geq  6 \text{ for n=3}$$
For $n < 3$ the minimum value occurs at $f'(z) = 0 \implies z = \sqrt{3n } < 3 \leq z$ so further more $f'(z \geq 3) > 0$ suggest that $f(z)$ is increasing. So, the min-value must occur at $z=3$ which gives the desired result.
