254. An interesting C. Mateescu's geometrical inequality.

by Virgil Nicula, Mar 13, 2011, 8:41 AM

Proposed problem. In any triangle $ABC$ have $\boxed{\sqrt{\frac {a}{b+c-a}}+\sqrt{\frac {b}{c+a-b}}+\sqrt{\frac {c}{a+b-c}}\ge\sqrt{\frac {5r+2R+s-3r\sqrt 3}r}}\ge$ $\sqrt{\frac {5r+2R}r}\ge 3$ .

Proof. This inequality is equivalently with the strong inequality $\boxed{AI+BI+CI\ge s+3r(2-\sqrt 3)}$ . Here appears its proof.

$\odot$ Using the relations $\boxed{\sqrt{\frac a{b+c-a}}=\sqrt{\frac{2R}r}\cdot\sin\frac A2}$ a.s.o., can transform the initial inequality in $\boxed{\sin\frac A2+\sin\frac B2+\sin\frac C2\ge\sqrt{\frac{5r+2R+s-3r\sqrt 3}{2R}}}\ (\ast)$ .

The inequality $\ (\ast)$ is truly in any triangle and it is stronger than the inequality $\boxed{\sin\frac A2+\sin\ \frac B2+\sin\frac C2\ge\frac 54+\frac r{2R}}\ (\ast\ast)$ which is truly only in an acute triangle.

The inegality $(\ast)$ is stronger than $(\ast\ast)$ and in the case of athe triangles for which $A\ge B\ge 60^{\circ}\ge C$ .

$\odot$ Using the relation $\boxed{AI=\sqrt{2Rr}\ \cdot\ \sqrt{\frac {b+c-a}a}}$ a.s.o. the inequality $\boxed{AI+BI+CI\ \ge\ s+3r(2-\sqrt 3)}$ becomes

$\boxed{\sqrt{\frac {b+c-a}a}+\sqrt{\frac {c+a-b}b}+\sqrt{\frac {a+b-c}c}\ge\frac{s+3r\cdot (2-\sqrt 3)}{\sqrt{2Rr}}}$ (Mateescu Constantin).


Proposed problem (V.N.). Prove that $\boxed{\sqrt{abc\left(a^3+b^3+c^3\right)}\ \le\ R\cdot (ab+bc+ca)}\ \le\ 4R(R+r)^2$ .

Maybe someone has an geometrical interpretation.

Proof. Firstly, we shall prove a very simple

Lemma.[ Let $a$ , $b$ , $c$ be the side-lengths of a triangle with circumradius $R$ .

If $x$ , $y$ , $z\in\mathbb{R}$ , then $:\ \boxed{\ yz\cdot a^2+zx\cdot b^2+xy\cdot c^2\ \le\ R^2\cdot\left(x+y+z\right)^2\ }\ \ (\ast)$ .

Proof.

Now, setting $x=bc$ , $y=ca$ and $z=ab$ in the inequality $(\ast)$ we obtain the proposed inequality :

$\begin{array}{cccc} abc(a^3+b^3+c^3)\ \le\ R^2\cdot (ab+bc+ca)^2\ \iff\ \boxed{\ \sqrt{abc(a^3+b^3+c^3)}\ \le\ R\cdot (ab+bc+ca)\ }\end{array}$ .

Remark. This very nice inequality has a beautiful geometrical interpretation as follows :

Denote the isotomic $J\ (bc:ca:ab)$ of the incenter $I\ (a:b:c)$ w.r.t. triangle $ABC$ . From the remarkable property -

The power $p_w(M)$ , where $M\ (x:y:z)$ w.r.t. the circumcircle $w$ of the triangle $ABC$ is given by the relation :

$ -p_{w}(M)=\frac{a^{2}yz+b^{2}zx+c^{2}xy}{(x+y+z)^{2}}=R^{2}-OM^{2}\le R^{2} $ - we obtain for $M:=J$ that :

$ \sum\left(a^{2}\cdot ca\cdot ab\right)\le (ab+bc+ca)^{2} R^{2} $ , which is in fact the proposed inequality .
This post has been edited 7 times. Last edited by Virgil Nicula, Nov 22, 2015, 10:28 AM

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