Attempt of a halfways nice solution. Problem. Let ABC be a triangle with $C\geq 60^{\circ}$. Prove the inequality $\left(a+b\right)\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 4+\frac{1}{\sin\frac{C}{2}}$. Solution. First, we equivalently transform the inequality in question: $\left(a+b\right)\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 4+\frac{1}{\sin\frac{C}{2}}$ $\Longleftrightarrow\ \ \ \ \ \left(a+b\right)\cdot\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{a+b}{c}\geq 4+\frac{1}{\sin\frac{C}{2}}$ $\Longleftrightarrow\ \ \ \ \ \left(a+b\right)\cdot\left(\frac{1}{a}+\frac{1}{b}\right)-4\geq\frac{1}{\sin\frac{C}{2}}-\frac{a+b}{c}$ $\Longleftrightarrow\ \ \ \ \ \frac{\left(a-b\right)^2}{ab}\geq\frac{1}{\sin\frac{C}{2}}-\frac{a+b}{c}$. Now, by the Mollweide formulas, $\frac{a+b}{c}=\frac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}}$ and $\frac{a-b}{c}=\frac{\sin\frac{A-B}{2}}{\cos\frac{C}{2}}$, so that $\frac{a-b}{a+b}=\frac{a-b}{c} : \frac{a+b}{c}=\frac{\sin\frac{A-B}{2}}{\cos\frac{C}{2}} : \frac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}}=\frac{\sin\frac{A-B}{2}\sin\frac{C}{2}}{\cos\frac{A-B}{2}\cos\frac{C}{2}}$. Now, $\cos^2\frac{C}{2}\leq 1$ (as the square of every cosine is $\leq 1$). On the other hand, the AM-GM inequality yields $ab\leq\frac14\left(a+b\right)^2$. Hence, $\frac{\left(a-b\right)^2}{ab}\geq\frac{\left(a-b\right)^2}{\frac14\left(a+b\right)^2}$ (since $ab\leq\frac14\left(a+b\right)^2$) $=4\left(\frac{a-b}{a+b}\right)^2=4\left(\frac{\sin\frac{A-B}{2}\sin\frac{C}{2}}{\cos\frac{A-B}{2}\cos\frac{C}{2}}\right)^2=\frac{4\sin^2\frac{A-B}{2}\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}\cos^2\frac{C}{2}}$ $\geq\frac{4\sin^2\frac{A-B}{2}\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}}$ (since $\cos^2\frac{C}{2}\leq 1$). $=\frac{4\left(2\sin\frac{A-B}{4}\cos\frac{A-B}{4}\right)^2\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}}=\frac{16\sin^2\frac{A-B}{4}\cos^2\frac{A-B}{4}\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}}$. Thus, instead of proving the inequality $\frac{\left(a-b\right)^2}{ab}\geq\frac{1}{\sin\frac{C}{2}}-\frac{a+b}{c}$, it will be enough to show the stronger inequality $\frac{16\sin^2\frac{A-B}{4}\cos^2\frac{A-B}{4}\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}}\geq\frac{1}{\sin\frac{C}{2}}-\frac{a+b}{c}$. Noting that $\frac{1}{\sin\frac{C}{2}}-\frac{a+b}{c}=\frac{1}{\sin\frac{C}{2}}-\frac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}}=\frac{1-\cos\frac{A-B}{2}}{\sin\frac{C}{2}}=\frac{2\sin^2\frac{A-B}{4}}{\sin\frac{C}{2}}$, we transform this inequality into $\frac{16\sin^2\frac{A-B}{4}\cos^2\frac{A-B}{4}\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}}\geq\frac{2\sin^2\frac{A-B}{4}}{\sin\frac{C}{2}}$, what, upon multiplication by $\frac{\cos^2\frac{A-B}{2}\sin\frac{C}{2}}{16\sin^2\frac{A-B}{4}}$ and rearrangement of terms, becomes $\sin^3\frac{C}{2}\cos^2\frac{A-B}{4}\geq\frac18\cos^2\frac{A-B}{2}$. But this trivially follows by multiplying the two inequalities $\sin^3\frac{C}{2}\geq\frac18$ (equivalent to $\sin\frac{C}{2}\geq\frac12$, what is true because $60^{\circ}\leq C\leq 180^{\circ}$ yields $30^{\circ}\leq\frac{C}{2}\leq 90^{\circ}$) and $\cos^2\frac{A-B}{4}\geq\cos^2\frac{A-B}{2}$ (follows from the obvious fact that $\left|\frac{A-B}{4}\right|\leq\left|\frac{A-B}{2}\right|$ since $\left|\frac{A-B}{2}\right|<90^{\circ}$, what is true because $\left|A-B\right|<180^{\circ}$, as the angles A and B, being angles of a triangle, lie between 0° and 180°). Hence, the problem is solved. Darij

Suppose w.l.o.g. $a\le b$. Because $C\ge 60^{\circ}$ there are only two cases: $\blacktriangleright\ 1^{\circ}.\ a\le b\le c$. In the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=72226 I offered a strengthening of this inequality: \[ \boxed {\ (a+b)\left(\frac 1a+\frac 1b+\frac 1c\right)\ge 4+\frac{1}{\sin \frac C2}\cdot \sqrt{1+\left(\frac{a-b}{2c}\right)^2}\ }\ge 4+\frac{1}{\sin \frac C2}\ . \] $\blacktriangleright\ 2^{\circ}.\ a\le c< b$. Proof. Remark two cases : $\boxed {\ \bullet\ 1.1.\ }\ a\le c\le \frac{a+b}{2}\le b\ .$ $\frac{(a-b)^2}{ab}+\frac{a+b}{c}\ge \frac{(a-b)^2}{ab}+2\ge 2\ge \frac{1}{\sin \frac C2}\Longrightarrow$$\frac{(a+b)^2}{ab}+\frac{a+b}{c}\ge 4+\frac{1}{\sin \frac C2}\ .$ $\boxed {\ \bullet\ 1.2.\ }\ a\le \frac{a+b}{2}<c<b\ .$ ??? I don't not know how can continue !. Now and here I beg a little help. I shall come back soon.