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.

Please help :)

  • $\begingroup$ Can you fix the formatting of (1) $\endgroup$ – joebloggs May 6 '14 at 16:17
  • $\begingroup$ @joebloggs Yeah, done! $\endgroup$ – Sawarnik May 6 '14 at 16:19
  • $\begingroup$ 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 $\endgroup$ – evil999man May 6 '14 at 16:42
  • $\begingroup$ @Awesome I know, but that was different. Here it is, math.stackexchange.com/questions/778373/… $\endgroup$ – Sawarnik May 6 '14 at 16:42
  • 1
    $\begingroup$ Good you asked it... I'll see if I can crack it... $\endgroup$ – 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.

  • 1
    $\begingroup$ (+1) but thats a brutal way of putting it down considering the fact that the inequality looks like a cute puppy ! ;) $\endgroup$ – r9m Dec 10 '14 at 20:40

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.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.