Prove $\frac{a^{2}}{b^{2}} +\frac{b^{2}}{c^{2}} +\frac{c^{2}}{a^{2}} +\frac{15abc}{4}\geq \frac{27}{4}$ for $a^{2}+b^{2}+c^{2}+abc=4$ 
Prove:
$$\frac{a^{2}}{b^{2}} +\frac{b^{2}}{c^{2}} +\frac{c^{2}}{a^{2}} +\frac{15abc}{4}\geq \frac{27}{4}$$ for $a,b,c>0$ such that $$a^{2}+b^{2}+c^{2}+abc=4.$$

I tried to note $a=\cos A$,... because of a known identity, I used $s=\frac{a^{2}}{b^{2}} +\frac{b^{2}}{c^{2}} +\frac{c^{2}}{a^{2}} \geq \frac{a}{b} +\frac{b}{c} +\frac{c}{a}$ and $s\geq \frac{a}{c} +\frac{b}{a} +\frac{c}{b}$, and then I added them. Another suggestion? Please!
I also noted $s=a+b+c,p=ab+bc+ac, r=abc$. The condion is equivalent with $s^2-2p+r=4$. Using the 2 inequalities which I mentioned above $\implies s\ge(a+b)(b+c)(a+c)/(2abc)-1$, and then is enought to prove that $2sp+15r^2\ge33r$, but I don't know if the last one is correct inequality. You can try to prove that. I'm desperate.

With the following link $a^2+b^2+c^2+abc=4 \Rightarrow abc \leq 1$ thus the maximum value is $1$ and we assume that $a=1,b=1,c=1$. $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{15abc}{4}$, we substitut $1$ we get exactly $\frac{27}{4}$, this is the maximum value that we could get. If $a,b,c$ is less that $1$ we could not reach $\frac{27}{4}$.So $\mathbf{a,b,c=1}$.
