The problem is to prove that for $a,b,c>0$ we have $$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{9abc}{4(a^3+b^3+c^3)}\geq \frac{15}{4}.$$
I have tried to use Bergstrom/Engel inequality to write, for example, $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2}$, and then to use Muirhead's inequalities to prove the remaining inequality - but unsuccessfully, so far...
The cyclically symmetric inequality is equivalent to: $$\color{red}{\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}-2 \right)} + \color{blue}{\left(\frac{b^2}{c^2}+\frac{c^2}{a^2}-\frac{b^2}{a^2}-1 \right)}+ \color{green}{\left(\frac{9abc}{4(a^3+b^3+c^3)}-\frac34\right)} \geqslant 0$$ $$\iff \color{red}{\frac{(a^2-b^2)^2}{a^2b^2}} + \color{blue}{\frac{(a^2-c^2)(b^2-c^2)}{a^2c^2}}+\color{green}{\frac{9abc-3(a^3+b^3+c^3)}{4(a^3+b^3+c^3)}}\geqslant 0$$ As $a^3+b^3+c^3-3abc=(a+b+c)((a-b)^2+(a-c)(b-c))$, we have above $\iff$ $$(a-b)^2\left(\color{red}{\frac{(a+b)^2}{a^2b^2}}-\color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}} \right)+(a-c)(b-c)\left(\color{blue}{\frac{(a+c)(b+c)}{a^2c^2}} - \color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}}\right)\geqslant 0$$
Now due to symmetry, we may assume $c=\min (a, b, c)$, hence it remains to show that under this condition, both $$\color{red}{\frac{(a+b)^2}{a^2b^2}}-\color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}} \geqslant 0, \qquad \color{blue}{\frac{(a+c)(b+c)}{a^2c^2}} - \color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}} \geqslant 0$$
However $(a^3+b^3)(a+b)\geqslant (a^2+b^2)^2\geqslant 4a^2b^2 \implies$ $$\color{red}{\frac{(a+b)^2}{a^2b^2}}\geqslant 4\frac{a+b}{a^3+b^3}> 4\frac{a+b}{a^3+b^3+c^3}\geqslant \frac83\frac{a+b+c}{a^3+b^3+c^3}> \color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}}$$ And $3(a^3+b^3+c^3)\geqslant (a+b+c)(a^2+b^2+c^2) \implies$ $$4(a^3+b^3+c^3)(a+c)(b+c)\geqslant \frac43(a+b+c)(a+c)(b+c)(a^2+b^2+c^2) \geqslant \frac43(a+b+c)(2c)(2c)(a^2)\geqslant 3(a+b+c)a^2c^2$$ $$\implies \color{blue}{\frac{(a+c)(b+c)}{a^2c^2}}\geqslant \color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}}$$ Hence the inequality holds true, with equality when $a=b=c$.
Another way.
By C-S $$\sum_{cyc}\frac{a^2}{b^2}=\sum_{cyc}\frac{a^4}{a^2b^2}\geq\frac{(a^2+b^2+c^2)^2}{a^2b^2+a^2c^2+b^2c^2}.$$ Thus, it's enough to prove that: $$\frac{(a^2+b^2+c^2)^2}{a^2b^2+a^2c^2+b^2c^2}+\frac{9abc}{4(a^3+b^3+c^3)}\geq\frac{15}{4}.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that $$\frac{(9u^2-6v^2)^2}{9v^4-6uw^3}+\frac{9w^3}{4(27u^3-27uv^2+3w^3)}\geq\frac{15}{4}$$ or $$\frac{(3u^2-2v^2)^2}{3v^4-2uw^3}+\frac{w^3}{4(9u^3-9uv^2+w^3)}\geq\frac{5}{4}$$ or $f(w^3)\geq0,$ where $$f(w^3)=8uw^6+(126u^4-138u^2v^2+4v^4)w^3+324u^7-756u^5v^2+441u^3v^4-9uv^8.$$ But by Schur $w^3\geq4uv^2-3u^3,$ which gives $$f'(w^3)=16uw^3+126u^4-138u^2v^2+4v^4\geq$$ $$\geq16u(4uv^2-3u^3)+126u^4-138u^2v^2+4v^4=78u^4-74u^2v^2+4v^4>0,$$ which says that $f$ increases.
Id est, it's enough to prove $f(w^3)\geq0$ for a minimal value of $w^3$, which happens in the following cases.
Let $c\rightarrow0^+$ and $b=1$.
Thus, we need to prove that $$\frac{(a^2+1)^2}{a^2}\geq\frac{15}{4},$$ which is true by AM-GM: $$\frac{(a^2+1)^2}{a^2}\geq\frac{(2a)^2}{a^2}=4>\frac{15}{4};$$ 2. Two variables are equal.
Let $b=c=1$.
Thus, we need to prove that: $$\frac{(a^2+2)^2}{2a^2+1}+\frac{9a}{4(a^3+2)}\geq\frac{15}{4}$$ or $$(a-1)^2(4a^5+8a^4-2a^3-4a^2+13a+2)\geq0$$ and we are done!
The following inequality is a bit of stronger.
Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{45abc}{4(a^3+b^3+c^3)}\geq\frac{27}{4}.$$
I will use the following inequality:
(Vasile Cirtoaje) If $x,y,z$ are positive real numbers, then:
$$(x+y+z)^3 \geq \frac{27}{4}(x^2y+y^2z+z^2x+xyz)$$
Setting $x=\frac{a^2}{b^2}, y = \frac{b^2}{c^2}, z = \frac{c^2}{a^2}$, we find
$$\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)^3\geq \frac{27}{4}\left(\frac{a^4}{b^2c^2}+\frac{b^4}{c^2a^2}+\frac{c^4}{a^2b^2}+1\right)$$
and using Cauchy-Schwarz
$$\frac{a^4}{b^2c^2}+\frac{b^4}{c^2a^2}+\frac{c^4}{a^2b^2} \geq \frac{(a^3+b^3+c^3)^2}{3a^2b^2c^2}$$
It remains to prove that:
$$\frac{27}{4}\left[\frac{(a^3+b^3+c^3)^2}{3a^2b^2c^2}+1\right]\geq \left[\frac{15}{4}-\frac{9abc}{4(a^3+b^3+c^3)}\right]^3$$
If we set
$$t=\frac{a^3+b^3+c^3}{abc}\geq 3$$
this becomes
$$\frac{27}{4}\left(\frac{t^2}{3}+3\right)\geq \left(\frac{15}{4}-\frac{9}{4t}\right)^3$$
Fully expanding this is equivalent with:
$$\frac{9}{64t^3}(t-3)(16t^4+48t^3-183t^2+126t-27)\geq 0$$
which is true as $t \geq 3$.
