Show this inequality $\dfrac{\sin{\frac{B}{2}}}{\sin{A}+\sin{B}}+\dfrac{\sin{\frac{C}{2}}}{\sin{A}+\sin{C}}\le\dfrac{1}{2\sin{A}}$ For any $\triangle ABC$, prove or disprove $$\dfrac{\sin{\frac{B}{2}}}{\sin{A}+\sin{B}}+\dfrac{\sin{\frac{C}{2}}}{\sin{A}+\sin{C}}\le\dfrac{1}{2\sin{A}}$$
by $$\sin{A}+\sin{B}=2\sin{\dfrac{A+B}{2}}\cos{\dfrac{A-B}{2}}$$
$$\sin{A}+\sin{C}=2\sin{\dfrac{A+C}{2}}\cos{\dfrac{A-C}{2}}$$
This leads to prove
$$\dfrac{\sin{\frac{B}{2}}}{\sin{\dfrac{A+B}{2}}\cos{\dfrac{A-B}{2}}}+\dfrac{\sin{\frac{C}{2}}}{\sin{\dfrac{A+C}{2}}\cos{\dfrac{A-C}{2}}}\le\dfrac{1}{\sin{A}}$$
 A: Let's assume $a,b,c$ are the sides of the triangle. We may suppose that $a=x+y$, $b=y+z$, and $c=x+z.$ If $p=\frac{a+b+c}{2}$, one can show that:
$$\sin \frac{B}{2}=\sqrt {\frac{(p-a)(p-c)}{ac}} =\sqrt{\frac{yz}{(x+y)(x+z)}}\\\sin \frac{C}{2}=\sqrt {\frac{(p-a)(p-b)}{ab}}=\sqrt{\frac{zx}{(x+y)(y+z)}}.$$
Moreover, $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$; hence, we just need to verify the relations below:
$$\frac{\sqrt{\frac{yz}{(x+y)(x+z)}}}{(x+y)+(y+z)}+\frac{\sqrt{\frac{zx}{(x+y)(y+z)}}}{(x+y)+(x+z)} \leq \frac{1}{2(x+y)} \\ \iff \frac{\sqrt {yz(x+y)}}{((x+y)+(y+z))\sqrt{x+z}}+\frac{\sqrt {zx(x+y)}}{((x+y)+(x+z))\sqrt {y+z}} \leq \frac{1}{2}.$$
On the other hand, we have:
$$\frac{\sqrt {yz(x+y)}}{((x+y)+(y+z))\sqrt{x+z}}+\frac{\sqrt {zx(x+y)}}{((x+y)+(x+z))\sqrt {y+z}} \\ \leq \frac{\sqrt {yz(x+y)}}{2\sqrt{(x+y)(y+z)}\sqrt {x+z}}+\frac{\sqrt {zx(x+y)}}{2\sqrt{(x+y)(x+z)}\sqrt{y+z}}=\frac{\sqrt {yz}+\sqrt{xz}}{2\sqrt{y+z}\sqrt{x+z}};$$
but:
$$\frac{\sqrt {yz}+\sqrt{xz}}{2\sqrt{y+z}\sqrt{x+z}} \leq \frac {1}{2};$$
because:
$$\sqrt {yz}+\sqrt{xz} \leq \sqrt{y+z}\sqrt{x+z} \\ \iff z(\sqrt x +\sqrt y)^2 \leq (y+z)(z+x)=z^2+zx+zy+xy \\ \iff 2z\sqrt {xy} \leq z^2+xy,$$
which is clear.
We are done.
