Let $p$ denote the semiperimeter of $\triangle ABC$ WLOG assume $a \geq b \geq c$ \begin{align*}\cos\frac{A-B}2 & =\cos\frac{A}2\cos\frac{B}2+\sin\frac{A}2\sin\frac{B}2 \\ & = \sqrt{\frac{p^2(p-a)(p-b)}{abc^2}} + \sqrt{\frac{(p-a)(p-b)(p-c)^2}{abc^2}}\end{align*} \[\implies \sum \cos\frac{A-B}2= \sum \frac{a+b}c \sin\frac{C}2\] Hence the inequality is equivalent to \[\sum \frac{a+b}c\sin\frac{C}2 \geq 3\sum \sin\frac{A}2-4\sum \sin^3\frac{A}2\] \[\iff \sum \frac{(b-c)(a-c)(a+b+c)}{abc} \sin\frac{C}2 \geq 0\] \[\iff \sum (c-b)(c-a) \sin\frac{C}2 \geq 0\] Due to assumption we have $(a-c)(b-c)\sin\frac{C}2 \geq 0$ So, it suffices to show that \[(b-a)(c-a)\sin\frac{A}2+(a-b)(c-b)\sin\frac{B}2 \geq 0\] \[\iff (a-b)\left((a-c)\sin\frac{A}2-(b-c)\sin\frac{B}2\right) \geq 0\] which is obviously true.

How may we order this book? Pierre.

Im asking the same too!!

After all, we are all Valentin's enthusiasts, hence Mathlinks. I would like to know how can I own a copy as well . Is it in English too?

It's only available in Romanian for now. Valentin is working on an extended edition in English, so I guess it's better to wait for the English version (which will become huge).

Who thinks i cannot read romanian??

i can also read romanian. is the book available online?

It's very nice solution - Valentin I want to see your book

We have : $\cos\frac{B-C}{2}-\sin\frac{3A}{2} = 2\sin\frac{A-B}{2}.\sin\frac{A-C}{2}$ So the ineq is equivalent to : $\sin\frac{A-B}{2}.\sin\frac{A-C}{2}+ \sin\frac{B-A}{2}.\sin\frac{B-C}{2} +\sin\frac{C-B}{2}.\sin\frac{C-A}{2} \geq 0$ (1) WLOG , suppose $0 <\frac{A}{2} <\frac{B}{2}<\frac{C}{2} < 90$ $(1) \iff \sin\frac{A-B}{2}.\sin\frac{A-C}{2}+\sin\frac{C-B}{2}[\sin\frac{C-A}{2}-\sin\frac{B-A}{2}] \geq 0$ , which is obviously true !

Equality holds when A = B = C = pi/3 then LHS = RHS = 3.

Let $\alpha=\frac{A}{2},\beta=\frac{B}{2},\gamma=\frac{C}{2}.$ Then we have \[ \sum_{\mathrm{cyclic}}\cos(\alpha-\beta)\geq\sum_{\mathrm{cyclic}}\sin3\alpha\;\;\;\Leftrightarrow\;\;\;\sum_{\mathrm{cyclic}}\cos^2\alpha+\sum_{\mathrm{cyclic}}\sin^2\beta\geq\sum_{\mathrm{cyclic}}\sin3\alpha. \] But $LHS=3$ and the inequality follows.

bodan wrote: Let $\alpha=\frac{A}{2},\beta=\frac{B}{2},\gamma=\frac{C}{2}.$ Then we have \[ \sum_{\mathrm{cyclic}}\cos(\alpha-\beta)\geq\sum_{\mathrm{cyclic}}\sin3\alpha\;\;\;\Leftrightarrow\;\;\;\sum_{\mathrm{cyclic}}\cos^2\alpha+\sum_{\mathrm{cyclic}}\sin^2\beta\geq\sum_{\mathrm{cyclic}}\sin3\alpha. \] But $LHS=3$ and the inequality follows. why are they equivalent ?

We have: $\sum\cos\frac{B-C}{2}\ge\sum\sin\frac{3A}{2}\\ \Longleftrightarrow 2\sum\sin\frac{B}{2}\sin\frac{C}{2}+\sum\sin\frac{A}{2}\ge 3\sum\sin\frac{A}{2}-4\sum\sin^{3}\frac{A}{2}\\ \iff 4\sum\sin^{3}\frac{A}{2}+2\sum\sin\frac{B}{2}\sin\frac{C}{2}\ge 2\sum\sin\frac{A}{2}\quad (*)$ Now, we'll prove that: $2\left(\sin^{3}\frac{A}{2}+\sin^{3}\frac{B}{2}\right)+\sin\frac{C}{2}\left(\sin\frac{A}{2}+\sin\frac{B}{2}\right)\ge\sin\frac{A}{2}+\sin\frac{B}{2}\quad (1)$ Indeed, we have: $(1)\iff 2\left(\sin^{2}\frac{A}{2}-\sin\frac{A}{2}\sin\frac{B}{2}+\sin^{2}\frac{B}{2}\right)+\sin\frac{C}{2}\ge 1\\ \iff 1-\cos A+1-\cos B+\cos\frac{A+B}{2}-\cos\frac{A-B}{2}+\sin\frac{C}{2}\ge 1\\ \iff 1+2\sin\frac{C}{2}\ge\cos A+\cos B+\cos\frac{A-B}{2},\;\textrm{what is truly.}$ Similar, we have: $2\left(\sin^{3}\frac{B}{2}+\sin^{3}\frac{C}{2}\right)+\sin\frac{A}{2}\left(\sin\frac{B}{2}+\sin\frac{C}{2}\right)\ge\sin\frac{B}{2}+\sin\frac{C}{2}\quad (2)$ $2\left(\sin^{3}\frac{C}{2}+\sin^{3}\frac{A}{2}\right)+\sin\frac{B}{2}\left(\sin\frac{C}{2}+\sin\frac{A}{2}\right)\ge\sin\frac{C}{2}+\sin\frac{A}{2}\quad (3)$ From $(1),\,(2)$ and $(3)$ we have $4\sum\sin^{3}\frac{A}{2}+2\sum\sin\frac{B}{2}\sin\frac{C}{2}\ge 2\sum\sin\frac{A}{2}\quad\mathbb{QED}$

can you by Schur inequality？

朱华伟老师please By schur inequality prove $ \sum\cos\frac{B-C}{2}\ge\sum\sin\frac{3A}{2}\\ \Longleftrightarrow 2\sum\sin\frac{B}{2}\sin\frac{C}{2}+\sum\sin\frac{A}{2}\ge 3\sum\sin\frac{A}{2}-4\sum\sin^{3}\frac{A}{2}\\ \iff 4\sum\sin^{3}\frac{A}{2}+2\sum\sin\frac{B}{2}\sin\frac{C}{2}\ge 2\sum\sin\frac{A}{2}\quad (*) $

Hmm I'm not entirely sure whether this approach works: Consider the function $f(x) = \sqrt{1-\cos x}$. Since $f''(x) = -\frac{\sin ^4 \frac{x}{2}}{(1-\cos x)^{\frac{3}{2}}}$, so $f(x)$ is concave down for all real $x \neq 2k\pi$. Using the identities $\sin \frac{\theta}{2} = \sqrt{\frac{1-\cos \theta}{2}}$ and $\cos \frac{\theta}{2} = \sqrt{\frac{1+\cos \theta}{2}}$, we rewrite the initial inequality as \begin{align*}\sqrt{1-\cos 3A}+\sqrt{1-\cos 3B}+\sqrt{1-\cos 3C} \le \sqrt{1+\cos (A-B)}+\sqrt{1+\cos (B-C)}+\sqrt{1+\cos(C-A)}.\end{align*} Since $ \cos \theta = -\cos (\theta+\pi)$, we have to prove that\begin{align*}\sqrt{1-\cos 3A}+\sqrt{1-\cos 3B}+\sqrt{1-\cos 3C} \le \sqrt{1-\cos (A-B+\pi)}+\sqrt{1-\cos (B-C+\pi)}+\sqrt{1-\cos(C-A+\pi)}.\end{align*} Without loss of generality let $\pi > A \ge B \ge C > 0$ (because $\cos \frac{A-B}{2} = \cos \frac{B-A}{2}$). Consider the sequences\begin{align*} \{x_i\}_{i=1}^{3} &= \left\{3A, 3B, 3C\right\}, \\ \{y_i\}_{i=1}^{3} &= \left\{B-C+\pi, C-A+\pi, A-B+\pi \right\}.\end{align*}We claim that $x_i$ majorizes $y_i$. First, $\sum_{i=1}^3 x_i = 3(A+B+C) = 3\pi$ and $\sum_{j=1}^3 y_j = 3\pi$. Now it is evident that $x_1 \ge y_1$ and $x_2 \ge y_2$, since \[3A-(B-C+\pi) = 3A-B+C-(A+B+C) = 2(A-B) \ge 0,\]\[3B-(C-A+\pi) = 3B-C+A-(A+B+C) = 2(B - C) \ge 0.\] Therefore $x_i$ majorizes $y_i$. By the Majorization Inequality (for concave down $f$), \[f(x_1) +f(x_2)+f(x_3) \le f(y_1)+f(y_2)+f(y_3)\] which implies our desired inequality.

Thank you very much, Robinpark for good solution! I like Karamata's inequality.

By the sum to product formulas, \begin{align*} \sum\limits_{\text{cyc}} \left[\cos \frac{B - C}{2} + \sin \frac{3A}{2}\right] &= \sum\limits_{\text{cyc}} \left[\sin\left(C + \frac{A}{2}\right)+ \sin \frac{3A}{2}\right] \\ &= \sum\limits_{\text{cyc}} 2\sin\left(\frac{C}{2} + A\right)\cos \frac{C - A}{2} \\ &\le \sum\limits_{\text{cyc}} 2\cos \frac{C - A}{2}, \end{align*}where the last step follows since $\cos \tfrac{C - A}{2} \ge 0$ and $\sin\left(\tfrac{C}{2} + A\right) \le 1.$ Simplification yields the desired result. $\square$