Proving a tough geometrical inequality, with equality in equilateral triangles. For any triangle with sides-lengths  $a$, $b$ and $c$ prove or disprove (1) and (2) :


*

*$$\sum_\mathrm{cyc} \frac{1}{\frac{(a+b)^2-c^2}{a^2}+1}\ge \frac34$$

*Equality in (1) holds if and only if the triangle is equilateral.


Playing with GeoGebra tells that they are correct, however, the proof eludes me.
Please help :)

Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, By C-S we obtain:
$$\sum_\mathrm{cyc} \frac{1}{\frac{(a+b)^2-c^2}{a^2}+1}=\sum_{cyc}\frac{a^2}{(a+b)^2-c^2+a^2}=$$
$$=\sum_{cyc}\frac{(y+z)^2}{y^2+5z^2+6yz+4zx}=\sum_{cyc}\frac{(y+z)^2(y+x)^2}{(y+x)^2(y^2+5z^2+6yz+4zx)}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}(y+z)(y+x)\right)^2}{\sum\limits_{cyc}(y+x)^2(y^2+5z^2+6yz+4zx)}=\frac{\sum\limits_{cyc}(x^4+6x^3y+6x^3z+11x^2y^2+24x^2yz)}{\sum\limits_{cyc}(x^4+6x^3y+6x^3z+11x^2y^2+40x^2yz)}.$$
Thus, it remains to prove that
$$4\sum\limits_{cyc}(x^4+6x^3y+6x^3z+11x^2y^2+24x^2yz)\geq3\sum\limits_{cyc}(x^4+6x^3y+6x^3z+11x^2y^2+40x^2yz)$$ or
$$\sum\limits_{cyc}(x^4+6x^3y+6x^3z+11x^2y^2-24x^2yz)\geq0,$$
which is obviously true.
Done!
