Triangles geometric inequalities

Question

Is there any technique for geometric inequalities for triangles? For example, I tried to use Ravi’s substitution

$a=x+y, b=y+z, c=z+x, s=x+y+z, s-a=z, s-b=x, s-c=y, A= \sqrt { xyz(x+y+z) }$

and

$r= \frac {A}{s} = \sqrt{\frac {xyz}{x+y+z}}),$

$R = \frac {abc} {4A}=\frac {(x+y)(y+z)(z+x) } {4 \sqrt { xyz(x+y+z) }}$

$\frac{r}{2R} = \frac {2xyz}{(x+y)(y+z)(z+x)} $

$r_a= \sqrt { \frac {xy}{z} (x+y+z)},$

to prove the inequality

$\sum_{cyc}^{} \frac {r_a}{a} \ge \sqrt {3(2+ \frac{r}{2R})}$

which becomes

$\sum_{cyc}^{} \sqrt {\frac {xy}{z} \frac{x+y+z}{x+y}} \ge \sqrt {3 (2+ \frac {2xyz}{(x+y)(y+z)(z+x) })}$

Then I tried to prove that

$\sin \frac{A}{2} + \sin \frac{B}{2}+ \sin \frac{C}{2} \le \sqrt {2 + \frac{r}{2R}} = \sqrt {2+ \frac {2xyz}{(x+y)(y+z)(z+x) }}$

Since

$\sin \frac{A}{2} + \sin \frac{B}{2}+ \sin \frac{C}{2} \le \frac {3}{2}$

then we have to prove

$\frac {3}{2} \le \sqrt {2+ \frac {2xyz}{(x+y)(y+z)(z+x) }}$

but I stuck. Thank you

Answer 1

A proof of $$\sin \frac{A}{2} + \sin \frac{B}{2}+ \sin \frac{C}{2} \le \sqrt {2 + \frac{r}{2R}}.$$

We need to prove that $$\sum_{cyc}\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}{2}}\leq\sqrt{2+\frac{\frac{2S}{a+b+c}}{\frac{abc}{2S}}}$$ or $$\sum_{cyc}\sqrt{\frac{(a+b-c)(a+c-b)}{4bc}}\leq\sqrt{2+\frac{16S^2}{4(a+b+c)abc}}$$ or

$$\sum_{cyc}\sqrt{\frac{(a+b-c)(a+c-b)}{bc}}\leq\sqrt{\frac{8abc+(a+b-c)(a+c-b)(b+c-a)}{abc}}$$ or $$\sum_{cyc}\sqrt{a(a+b-c)(a+c-b)}\leq\sqrt{8abc+(a+b-c)(a+c-b)(b+c-a)}$$ or $$\sum_{cyc}\sqrt{(y+z)4yz}\leq\sqrt{8(x+y)(x+z)(y+z)+8xyz}$$ or $$\sqrt{2(x+y+z)(xy+xz+yz)}\geq\sum_{cyc}\sqrt{xy(x+y)}$$ or $$2\sum_{cyc}(x^2y+x^2z+xyz)\geq\sum_{cyc}(x^2y+x^2z+2\sqrt{xyxz(x+y)(x+z)})$$ or $$\sum_{cyc}(x^2y+x^2z+2xyz)\geq2\sum_{cyc}x\sqrt{yz(x+y)(x+z)},$$ which is true by AM-GM: $$2\sum_{cyc}x\sqrt{yz(x+y)(x+z)}\leq\sum_{cyc}x(y(x+z)+z(x+y))=\sum_{cyc}(x^2y+x^2z+2xyz).$$ Done!

A proof of $$\sum_{cyc}^{} \frac {r_a}{a} \ge \sqrt {3(2+ \frac{r}{2R})}.$$ We need to prove that $$\sum_{cyc}\frac{2S}{a(b+c-a)}\geq\sqrt{\frac{3(8abc+(a+b-c)(a+c-b)(b+c-a))}{4abc}}$$ or $$\sum_{cyc}\frac{2\sqrt{(x+y+z)xyz}}{2x(y+z)}\geq\sqrt{\frac{3(8(x+y)(x+z)(y+z)+8xyz)}{4(x+y)(x+z)(y+z)}}$$ or $$\sum_{cyc}yz(x+y)(x+z)\geq\sqrt{6(xy+xz+yz)xyz(x+y)(x+z)(y+z)}.$$ Now, let $xy=r$, $xz=q$ and $yz=p$.

Thus, we need to prove that $$\sum_{cyc}(p+q)(p+r)\geq\sqrt{6(p+q+r)(p+q)(p+r)(q+r)}$$ or $$\sum_{sum}(p^4-p^2q^2)\geq0$$ or $$\sum_{cyc}(p^2-q^2)^2\geq0.$$ Done!