232. Some geometrical inequalities (own).

by Virgil Nicula, Feb 28, 2011, 5:49 PM

Proposed problem 1. Prove that $\triangle ABC\implies 4(a + b +$ $ c)\ \le\boxed{\ \sum \frac {(b + c)^2}{b + c - a}\ \le\ 4(a + b + c) + \frac {2(4R + r)(R - 2r)}{r}\cdot\sqrt {\frac {4R + r}{16R - 5r}}\ }$ .

Proof. I"ll use the remarkable identity $\boxed{\ \sum\ \frac {(b+c)^2}{b+c-a}=8s+\frac {R-2r}r\cdot 2s\ }$ (its proof is left to the interest reader). Thus, our inequality is equivalent to

$8s+\frac {R-2r}r\cdot 2s\ \le\ 8s$ $+\frac {2(4R+r)(R-2r)}r\cdot\sqrt{\frac {4R+r}{16R-5r}}\iff \boxed{\ s\ \le\ 4R+r\cdot\sqrt{\frac {4R+r}{16R-5r}}\ }$ . Comparing it to Gerretsen's inequality

$\boxed{\ s^2\ \le\ 4R^2+4Rr+3r^2\ }$ it remains to show that $\boxed{\ (4R^{2}+4Rr+3r^{2})(16R-5r)\ \le\ (4R+r)^{3}\ }$ $\iff$

$64R^{3}+44R^{2}r+28Rr^{2}-15r^{3}\le 64R^{3}+48R^{2}r+12Rr^{2}+r^{3}\iff 4r(R-2r)^{2}\ge 0 $ , which is true.


Proposed problem 2. Prove that $\triangle ABC\implies\frac {a}{b+c-a}+\frac {b}{c+a-b}+\frac {c}{a+b-c}\ \ge\ \frac {b+c-a}{a}$ $ +\frac {c+a-b}{b}+\frac {a+b-c}{c}+\frac {R-2r}{R}\cdot\frac {8R-r}{4(2R-r)}$ .

Proof. We will use the following set of identities $\left\|\begin{aligned}\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}&=\frac {s^2+r^2-2Rr}{2Rr};\\\frac a{b+c-a}+\frac b{c+a-b}+\frac c{a+b-c}&=\frac{2R-r}{r};\end{aligned}\right\|$ . Substituting all these values

it is enough to check that $\frac{2R-r}{r}\geq \frac{s^2+r^2-2Rr}{2Rr}-3+\frac{R-2r}{R}\cdot \frac{8R-r}{4(2R-r)}$ . Using the fact that $s^2 \le \frac{R(4R+r)^2}{2(2R-r)}$ , we have to show that

$\frac{2R}{r}\geq \frac{\frac{R(4R+r)^2}{2(2R-r)}+r^2}{2Rr}-3-\frac{R-2r}{R}\cdot\frac{8R-r}{4(2R-r)}$ . Let $t=\frac{R}{2r}\geq 1$ we have to show that , $4t\geq \frac{\frac{t(8t+1)^2}{4t-1}+1}{4t}-3-\left(1-\frac 1t\right)\left(\frac{16t-1}{4(4t-1)}\right)$ .

This, however, factorises into $\frac{(t-1)(16t-1)}{2t(4t-1)}\geq 0;$ which is obviously true.


Proposed problem 3. $\boxed{\ \mathrm{Prove\ that}\ \triangle ABC\implies\frac {1}{a(b+c-a)}+\frac {1}{b(c+a-b)}+\frac {1}{c(a+b-c)}\ \ge\ \frac {5R-2r}{8R^2r}\ }$

Proof.
This post has been edited 18 times. Last edited by Virgil Nicula, Apr 20, 2018, 3:22 PM

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