# 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) :

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

• Can you fix the formatting of (1) – joebloggs May 6 '14 at 16:17
• @joebloggs Yeah, done! – Sawarnik May 6 '14 at 16:19
• There is a same question here that used just the half angle identity in that term in denominator. I can't seem to find it – evil999man May 6 '14 at 16:42
• @Awesome I know, but that was different. Here it is, math.stackexchange.com/questions/778373/… – Sawarnik May 6 '14 at 16:42
• Good you asked it... I'll see if I can crack it... – evil999man May 6 '14 at 16:45

We can set $a=y+z,b=x+z,c=x+y$ and find the minimum of $$H(x,y,z)=\sum_{cyc}\frac{(x+z)^2}{5x^2+4xy+6xz+z^2}$$ over ${\mathbb{R}^+}^3$. Since cyclic permutations of the variables are allowed, we can assume without loss of generality that $a\leq b\leq c$ or $a\geq b\geq c$, that is equivalent to say that $y$ always lie between $x$ and $z$. Since the starting inequality is homogeneous, we can assume also that $x+y+z=3$. As a sum of non-negative convex functions, $H$ is a non-negative convex function too, hence it has a unique minimum point over $D={\mathbb{R}^+}^3\cap\{(x,y,z): x+y+z=3\}$. By computing the partial derivatives of $H$ we can see that $(1,1,1)$ is a stationary point inside the domain $D$, hence it is a minimum (by convexity, maxima lie on the boundary and there are no saddle points). This gives: $$H(x,y,z)\geq H(1,1,1) = \frac{3}{4}$$ as wanted.
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.