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.
 A: 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 $\mathbf{a,b,c=1}$.
A: This is  problem 28003 published by Marius St\u{a}nean in [Problem 28003, Gazeta matematic\u{a}, Seria B, 126 (2021), no. 2].
Its solution is published in volume 126, no. 9. Marian Cucoane\c{s}
showed in [Asupra problemei 28003 din Gazeta Matematic\u{a} nr. 2/2021, Gazeta matematic\u{a}, Seria B, 127 (2022), no. 2, 60--61] that the inequality
$$
 \frac{a^2}{b^2}+ \frac{b^2}{c^2} + \frac{c^2}{a^2} +kabc \ge k+3      \tag{*}\label{*}
$$
holds for all $k\le \frac{9+2\sqrt{14}}{3}$. An even stronger version
is due to Titu Zvonaru, who proved in  Din nou despre problema 28003 din G.M.--B 2/2021, Gazeta matematic\u{a}, Seria B, 127 (2022), no. 10, 437--438  that the largest $k$
for which inequality~\eqref{*} is true for positive numbers satisfying
$a^2+b^2+c^2+abc=4$ is greater than or equal to $22/3$ and less than $8.2$.
His proof is rather computational. After deconditioning  by using the well-known transformation
$$
 a=\frac{2\sqrt{yz}}{\sqrt{(y+x)(z+x)}}, \ldots 
$$
and ordering the nonnegative variables $x$, $y$, $z$, one reduces the problem to showing that a homogeneous polynomial of degree six  is nonnegative when its arguments are so. This is achieved by pointing out four obvious inequalities whose addition gives the desired one.
