$a$, $b$ and $c$ are the side of a triangle $\triangle ABC$ that is not obtuse and let $R$ be it the circumcenter. Prove that
$$\frac{\left(\sum_{cyc} a^2\right)^2}{\prod_{cyc}(a^2+b^2)}<\frac{5}{2R^2}$$
My attempt is to write everything in terms of $R$, $r$ and $s$ and manage to use something like Blundon’s inequality. The problem is I didn't find a simple formula for $\prod_{cyc}(a^2+b^2)$.
Let $a^2+b^2-c^2=z$, $a^2+c^2-b^2=y$ and $b^2+c^2-a^2=x$.
Hence, $a^2=\frac{y+z}{2}$, $b^2=\frac{x+z}{2}$, $c^2=\frac{x+y}{2}$ and we need to prove that: $$\frac{8(x+y+z)^2}{\prod\limits_{cyc}(2x+y+z)}<\frac{5}{2\cdot\frac{a^2b^2c^2}{16S^2}}$$ or $$\frac{8(x+y+z)^2}{\prod\limits_{cyc}(2x+y+z)}<\frac{5(xy+xz+yz)}{2\cdot\frac{\prod\limits_{cyc}(x+y)}{8}}$$ or $$2\prod_{cyc}(x+y)(x+y+z)^2<5(xy+xz+yz)\prod_{cyc}(2x+y+z),$$ which is obvious and very very not strong.
By the way, the inequality $$\frac{(a^2+b^2+c^2)^2}{(a^2+b^2)(a^2+c^2)(b^2+c^2)}<\frac{1}{2R^2}$$ gives $$2\prod_{cyc}(x+y)(x+y+z)^2<(xy+xz+yz)\prod_{cyc}(2x+y+z),$$ which is stronger and also obvious because it's just $$\sum_{sym}(x^3y^2+2x^3yz+5x^2y^2z)>0$$