245. An inequality in an acute triangle.

by Virgil Nicula, Mar 5, 2011, 12:39 PM

$\bigodot$ Prove that in an acute triangle $ ABC$ with circumradius $ R$ and inradius $ r$ have $ \boxed {\ \sum \frac {a^2}{b^2 + c^2 - a^2}\ \ge\ \left(\frac Rr\right)^2 - 1\ }\ \ge\ \frac {(a + b + c)^2}{a^2 + b^2 + c^2}$ .

Proof. Prove easily that the identities $ \left\|\begin{array}{cc} (1) & 4S = \left(b^2 + c^2 - a^2\right)\cdot\tan A \\
 \\
(2) & \sum\sin A\cos A = \frac {S}{R^2}\end{array}\right\|$ and the inequality $ \sum\tan A\ge\frac {S}{r^2}\ (3)$ .

Therefore , $ \sum \frac {a^2}{b^2 + c^2 - a^2}\ \stackrel{(1)}{ = }\ \frac {1}{4S}\cdot \sum a^2\tan A =$ $ \frac {R^2}{S}\cdot\sum\frac {\sin^3 A}{\cos A} =$ $ \frac {R^2}{S}\cdot\sum\frac {\sin A(1 - \cos^2A)}{\cos A} =$

$ \frac {R^2}{S}\cdot\left(\sum\tan A - \sum\sin A\cos A\right)\ \stackrel{(2)\wedge (3)}{\ge}\ \frac {R^2}{S}\cdot\left(\frac {S}{r^2} - \frac {S}{R^2}\right) =$ $ \left(\frac Rr\right)^2 - 1\ \ \implies\ \ \sum \frac {a^2}{b^2 + c^2 - a^2}\ge\left(\frac Rr\right)^2 - 1$ .

Remark. $ \sum \frac {a^2}{b^2 + c^2 - a^2}\ \stackrel{C.B.S.}{\ge}\ \frac {\left(\sum a\right)^2}{\sum \left(b^2 + c^2 - a^2\right)} = \frac {(a + b + c)^2}{a^2 + b^2 + c^2}$ $ \implies$ $ \sum \frac {a^2}{b^2 + c^2 - a^2}\ \ge\ \frac {(a + b + c)^2}{a^2 + b^2 + c^2}$

and $ \left(\frac Rr\right)^2 - 1\ge 3\ge\frac {(a + b + c)^2}{a^2 + b^2 + c^2}\ \implies\ \left(\frac Rr\right)^2 - 1\ \ge\ \frac {(a + b + c)^2}{a^2 + b^2 + c^2}$ .


Remark (Pikachu). For the another inequality can be made stronger : $ \frac {R^2}{r^2} - 1\ \geq\ \frac {9(a^2 + b^2 + c^2)}{(a + b + c)^2}\ \ (*)$ .

Proof. Nice your stronger inequality ! Thus, $ \frac {R^2}{r^2} - 1\ \geq\ \frac {9(a^2 + b^2 + c^2)}{(a + b + c)^2}$ $ \Longleftrightarrow 9r^2(s^2 - r^2 - 4Rr)\le 2s^2(R^2 - r^2)$ .

From the Gerretsen's inequality in any triangle obtain $ 9r^2(s^2 - r^2 - 4Rr)\le 9r^2(4R^2 + 2r^2)\equiv U$ . From the Walker's

inequality
in an acute triangle obtain $ V\equiv 2(2R^2 + 8Rr + 3r^2)(R^2 - r^2)\le 2s^2(R^2 - r^2)$ . I"ll prove that $ U\le V$ , i.e.

$ 9r^2(4R^2 + 2r^2)\le 2(2R^2 + 8Rr + 3r^2)(R^2 - r^2)\ \ (1)$ . Indeed, if denote $ \frac Rr = t\ge 2$ , then the relation $ (1)$ becomes

$ 9(2t^2 + 1)\le (2t^2 + 8t + 3)(t^2 - 1)\Longleftrightarrow$ $ 2t^4 + 8t^3 - 17 - 8t - 12\ge 0$ $ \Longleftrightarrow$ $ (t - 2)(2t^3 + 12t^2 + 7t + 6)\ge 0$ ,

what is truly because $ t\ge 2$ . In conclusion, $ 9r^2(s^2 - r^2 - 4Rr)\le U\le V\le 2s^2(R^2 - r^2)\ \implies$ required inequality $ (*)$ .

Remark. Therefore, there is the chain $ \boxed {\ \sum \frac {a^2}{b^2 + c^2 - a^2}\ \ge\ \left(\frac Rr\right)^2 - 1\ \ge\ \frac {9(a^2 + b^2 + c^2)}{(a + b + c)^2}\ \ge\ \frac {(a + b + c)^2}{a^2 + b^2 + c^2}\ }$ .

Prove easily that the inequality $ (*)$ is equivalently with nice inequality $ \frac {(a - b)^2 + (b - c)^2 + (c - a)^2}{a^2 + b^2 + c^2}\ \le\ \frac {3(R^2 - 4r^2)}{R^2 - r^2}$ .


Remark (Mateescu Constantin). It also holds the following stronger inequality in an acute-angled triangle $ABC\ :$

$\bullet\ \boxed{\ \sum\ \frac{a^2}{b^2+c^2-a^2}\ \ge\ 2\cdot\left(\frac Rr\right)^2-\frac{3R^2\sqrt 3}{S}-1\ }\ \ge\ \left(\frac Rr\right)^2-1$ .
This post has been edited 3 times. Last edited by Virgil Nicula, Nov 22, 2015, 12:23 PM

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