Triangle inequality with the exradii $r_{a}$, $r_{b}$, $r_{c}$, the medians $m_{a}$, $m_{b}$, $m_{c}$

Question

Given a triangle with the exradii $r_{a}$, $r_{b}$, $r_{c}$, the medians $m_{a}$, $m_{b}$, $m_{c}$. Show that $$r_{a}^{2}+ r_{b}^{2}+ r_{c}^{2}\geq 3\sqrt{3}. S+ \left ( m_{a}- m_{b} \right )^{2}+ \left ( m_{b}- m_{c} \right )^{2}+ \left ( m_{c}- m_{a} \right )^{2}$$ I can only show the weaker inequality $$r_{a}^{2}+ r_{b}^{2}+ r_{c}^{2}\geq 3\sqrt{3}. S$$ I need to the help. Thanks!

Answer 1

Let $a=y+z$, $b=x+z$ and $c=x+y$.

Hence, $x$, $y$ and $z$ they are positives and we need to prove that $$\sum_{cyc}\frac{4S^2}{(b+c-a)^2}\geq3\sqrt{3xyz(x+y+z)}+2\sum_{cyc}(m_a^2-m_am_b)$$ or $$\sum_{cyc}\frac{xyz(x+y+z)}{x^2}+\frac{1}{2}\sum_{cyc}\sqrt{\left(4x(x+y+z)+(y-z)^2\right)\left(4y(x+y+z)+(x-z)^2\right)}\geq3\sqrt{3xyz(x+y+z)}+3\sum_{cyc}(x^2+xy).$$ Now, $$\sqrt{3xyz(x+y+z)}\leq xy+xz+yz$$ and by C-S and AM-GM \begin{align} &\sum_{cyc}\sqrt{\left(4x(x+y+z)+(y-z)^2\right)\left(4y(x+y+z)+(x-z)^2\right)} \\ =&\sum_{cyc}\sqrt{(4x^2+y^2+z^2+4xy+4xz-2yz)(x^2+4y^2+4xy+4yz-2xz)} \\ =&\sum_{cyc}\sqrt{((2x+y-z)^2+8xz)((2y+x-z)^2+8yz)} \\ \geq&\sum_{cyc}((2x+y-z)(2y+x-z)+8z\sqrt{xy}) \\ \geq&\sum_{cyc}(2x+y-z)(2y+x-z)+24\sqrt[3]{x^2y^2z^2} \\ =&\sum_{cyc}(5x^2-xy)+24\sqrt[3]{x^2y^2z^2}. \end{align} Thus, it's enough to prove that $$\sum_{cyc}\frac{yz(x+y+z)}{x}+\frac{1}{2}\sum_{cyc}(5x^2-xy)+12\sqrt[3]{x^2y^2z^2}\geq3(xy+xz+yz)+3\sum_{cyc}(x^2+xy)$$ or $$\frac{2(x^2y^2+x^2z^2+y^2z^2)(x+y+z)}{xyz}+24\sqrt[3]{x^2y^2z^2}\geq\sum_{cyc}(x^2+13xy).$$ Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Thus, we need to prove that $$\frac{2(9v^4-6uw^3)3u}{w^3}+24w^2\geq9u^2+33v^2$$ or $f(v^2)\geq0,$ where $$f(v^2)=\frac{2u(9v^4-6uw^3)}{w^3}+8w^2-3u^2-11v^2.$$ But, $$f'(v^2)=\frac{36uv^2}{w^3}-11>0,$$ which says that it's enough to prove that $f(v^2)\geq0$ for the minimal value of $v^2$,

which happens for equality case of two variables.

Since the last inequality is homogeneous, then after replacing $x$ on $x^3$ we need to prove that $$\frac{2(2x^6+1)(x^3+2)}{x^3}+24x^2\geq x^6+2+26x^3+13$$ or $$(x-1)^2(3x^7+6x^6+9x^5-6x^4+3x^3+12x^2+8x+4)\geq0,$$ which is obvious.

Answer 2

There is another proof bases on $R, r, s$ relationship as follows:

Let $R, r, s$ be the circumcircle radius, incircle radius, and semi-perimeter of the triangle. Then $a, b, c, r_a, r_b, r_c$ satisfy the following relationship: $$ R\ge 2r$$ $$ r_a+r_b+r_c=4R+r $$ $$r_ar_b+r_br_c+r_cr_a=s^2 $$ $$m_a^2+m_b^2+m_c^2=\dfrac{3}{4}(a^2+b^2+c^2) $$ $$ a^2+b^2+c^2=2(s^2-4Rr-r^2)$$ $$ s^2\le 4R^2+4Rr+3r^2\ \text{(Gerretsen inequality)}$$

We have: $R\ge 2r$ then: $4R^2\ge 16r^2$. Thus:

$$32R^2+28Rr+5r^2\ge 28R^2+28Rr+21r^2$$

By Gerretsen inequality we have: $32R^2+28Rr+5r^2\ge 7s^2$.

Hence: $32R^2+16Rr+2r^2+12Rr+3r^2\ge7s^2$ or equivalently: $$2(4R+r)^2+12Rr+3r^2\ge 7s^2$$ $$\Leftrightarrow 2(4R+r)^2-4s^2\ge 3s^2-12Rr-3r^2$$ $$\Leftrightarrow 2\Big((4R+r)^2-2s^2\Big)\ge 3(s^2-4Rr-r^2)$$ $$\Leftrightarrow (4R+r)^2-2s^2\ge \dfrac{3}{2}(s^2-4Rr-r^2)$$ $$ \Leftrightarrow r_a^2+r_b^2+r_c^2\ge 2\cdot\dfrac{3}{4}(a^2+b^2+c^2)=m_a^2+m_b^2+m_c^2$$

Now, note that $m_a, m_b, m_c$ are also the sidelength of a triangle, with area equals to $\dfrac{3}{4}S$, then by the well-known Finsler-Hadwiger inequality we have: $$ m_a^2+m_b^2+m_c^2 \ge 3\sqrt{3}S+(m_a-m_b)^2+(m_b-m_c)^2+(m_c-m_a)^2 $$

Combine the two results above we obtain: $$ r_a^2+r_b^2+r_c^2 \ge 3\sqrt{3}S+(m_a-m_b)^2+(m_b-m_c)^2+(m_c-m_a)^2 $$