Prove this trigonometric inequality about the angles of $\triangle ABC$

Question

In $\Delta ABC$ show that $$\cos{\frac{A}{2}}+\cos\frac{B}{2}+\cos\frac{C}{2}\ge \frac{\sqrt{3}}{2} \left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\frac{A-B}{2}\right)$$

since $$\frac{\sqrt{3}}{2}\left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\dfrac{A-B}{2}\right)=\frac{\sqrt{3}}{2}\sum\cos\frac{A}{2}\cos\frac{B}{2}+\sin\frac{B}{2} \sin\frac{C}{2}$$

Answer 1

Let $a=y+z$, $b=x+z$ and $c=x+y$.

Hence, $x$, $y$ and $z$ are positives and we need to prove that $$\sum_{cyc}\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2bc}}{2}}\geq$$ $$\geq\frac{\sqrt{3}}{2}\sum_{cyc}\left(\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2bc}}{2}\cdot\frac{1+\frac{a^2+c^2-b^2}{2ac}}{2}}+\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}{2}\cdot\frac{1-\frac{a^2+c^2-b^2}{2ac}}{2}}\right)$$ or $$\sum_{cyc}\sqrt{\frac{(a+b+c)(b+c-a)}{bc}}\geq$$ $$\geq\frac{\sqrt3}{4c\sqrt{ab}}\sum_{cyc}(a+b+c+a+b-c)\sqrt{(b+c-a)(a+c-b)}$$ or $$2\sum_{cyc}\sqrt{c^2ab(a+b+c)(a+b-c)}\geq\sqrt3\sum_{cyc}(a+b)\sqrt{ab(a+c-b)(b+c-a)}$$ or $$2\sum_{cyc}\sqrt{(x+y)^2(x+z)(y+z)(x+y+z)z}\geq\sqrt3\sum_{cyc}(x+y+2z)\sqrt{xy(x+z)(y+z)}$$ or $$\frac{4}{\sqrt3}\sqrt{(x+y+z)\prod_{cyc}(x+y)}\sum_{cyc}\sqrt{z(x+y)}+$$ $$+\sum_{cyc}(x+y+2z)\left(x(y+z)+y(x+z)-2\sqrt{xy(x+z)(y+z)}\right)\geq$$ $$\geq\sum_{cyc}(x+y+2z)(x(y+z)+y(x+z))$$ or $$\frac{4}{\sqrt3}\sqrt{(x+y+z)\prod_{cyc}(x+y)}\sum_{cyc}\sqrt{z(x+y)}+$$ $$+\sum_{cyc}(x+y+2z)\left(\sqrt{x(y+z)}-\sqrt{y(x+z)}\right)^2\geq$$ $$\geq\sum_{cyc}(5x^2y+5x^2z+6xyz)$$ or $$\frac{4}{\sqrt3}\sqrt{(x+y+z)\prod_{cyc}(x+y)}\sum_{cyc}\sqrt{z(x+y)}+$$ $$+\sum_{cyc}\frac{(x+y+2z)z^2(x-y)^2}{\left(\sqrt{x(y+z)}+\sqrt{y(x+z)}\right)^2}\geq\sum_{cyc}(5x^2y+5x^2z+6xyz).$$ Now, since by Holder $$\sum_{cyc}\sqrt{z(x+y)}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{z(x+y)}\right)^2\sum\limits_{cyc}z^2(x+y)^2}{\sum\limits_{cyc}z^2(x+y)^2}}\geq$$ $$\geq\sqrt{\frac{8(xy+xz+yz)^3}{2\sum\limits_{cyc}(x^2y^2+x^2yz)}}=2\sqrt{\frac{(xy+xz+yz)^3}{\sum\limits_{cyc}(x^2y^2+x^2yz)}}$$ and by C-S $$\left(\sqrt{x(y+z)}+\sqrt{y(x+z)}\right)^2\leq(1+1)(x(y+z)+y(x+z))=2(2xy+xz+yz),$$ it remains to prove that $$\frac{8}{\sqrt3}\sqrt{\frac{(x+y+z)\prod\limits_{cyc}(x+y)(xy+xz+yz)^3}{\sum\limits_{cyc}(x^2y^2+x^2yz)}}+\frac{1}{2}\sum_{cyc}\frac{(x-y)^2z^2(x+y+2z)}{2xy+xz+yz}\geq$$ $$\geq\sum_{cyc}(5x^2y+5x^2z+6xyz).$$ We'll prove that $$\sum_{cyc}\frac{(x-y)^2z^2(x+y+2z)}{2xy+xz+yz}\geq\frac{2(x+y+z)\sum\limits_{cyc}z^2(x-y)^2}{3(xy+xz+yz)}.$$ Indeed, let $x\geq y\geq z$.

Hence, $$\sum_{cyc}\frac{(x-y)^2z^2(x+y+2z)}{2xy+xz+yz}-\frac{2(x+y+z)\sum\limits_{cyc}z^2(x-y)^2}{3(xy+xz+yz)}=$$ $$=\frac{\sum\limits_{cyc}z^2(x-y)^2(3(x+y+2z)(xy+xz+yz)-2(x+y+z)(2xy+xz+yz))}{3(2xy+xz+yz)(xy+xz+yz)}=$$ $$=\frac{\sum\limits_{cyc}z^2(x-y)^2(4z^2x+4z^2y+x^2z+y^2z+4xyz-x^2y-y^2x)}{3(2xy+xz+yz)(xy+xz+yz)}\geq$$ $$\geq\frac{\sum\limits_{cyc}z^2(x-y)^2(z^2x+z^2y+x^2z+y^2z-x^2y-y^2x)}{3(2xy+xz+yz)(xy+xz+yz)}\geq$$ $$\geq\frac{z^2(x-y)^2(z^2x+z^2y+x^2z+y^2z-x^2y-y^2x)}{(2xy+xz+yz)(xy+xz+yz)}+$$ $$+\frac{y^2(x-z)^2(y^2x+y^2z+x^2y+z^2y-x^2z-z^2x)}{3(2xz+xy+yz)(xy+xz+yz)}\geq$$ $$\geq\frac{z^2(x-y)^2(z^2x+x^2z-x^2y-y^2x)}{3(2xy+xz+yz)(xy+xz+yz)}+$$ $$+\frac{y^2(x-y)^2(y^2x+x^2y-x^2z-z^2x)}{3(2xz+xy+yz)(xy+xz+yz)}=$$ $$=\frac{x(x+y+z)(x-y)^2(y-z)}{3(xy+xz+yz)}\left(\frac{y^2}{2xz+xy+yz}-\frac{z^2}{2xy+xz+yz}\right)\geq0.$$ Id est, it's enough to prove that $$\frac{8}{\sqrt3}\sqrt{\frac{(x+y+z)\prod\limits_{cyc}(x+y)(xy+xz+yz)^3}{\sum\limits_{cyc}(x^2y^2+x^2yz)}}+\frac{(x+y+z)\sum\limits_{cyc}z^2(x-y)^2}{3(xy+xz+yz)}\geq$$ $$\geq\sum_{cyc}(5x^2y+5x^2z+6xyz).$$ Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v>0$, and $xyz=w^3$.

Hence, we need to prove that $$\frac{8}{\sqrt3}\sqrt{\frac{(x+y+z)\prod\limits_{cyc}(x+y)(xy+xz+yz)^3}{\sum\limits_{cyc}(x^2y^2+x^2yz)}}+\frac{2(x+y+z)\sum\limits_{cyc}(x^2y^2-x^2yz)}{3(xy+xz+yz)}\geq$$ $$\geq\sum_{cyc}(5x^2y+5x^2z+6xyz)$$ or $$\frac{8}{\sqrt3}\sqrt{\frac{3u(9uv^2-w^3)\cdot27v^6}{9v^4-6uw^3+3uw^3}}+\frac{2\cdot3u(9v^4-9uw^3)}{9v^2}\geq5(9uv^2-3w^3)+18w^3$$ or $$8\sqrt{\frac{uv^6(9uv^2-w^3)}{3v^4-uw^3}}\geq\frac{9uv^4+6u^2w^3+v^2w^3}{v^2}$$ or $f(w^3)\geq0$, where $$f(w^3)=333u^2v^{12}-243u^3v^8w^3-118uv^{10}w^3-3v^8w^6-18u^2v^6w^6+$$ $$+36u^5w^9+12u^3v^2w^9+uv^4w^9.$$ We see that $$f'(w^3)=-243u^3v^8-118uv^{10}-6v^8w^3-36u^2v^6w^3+108u^5w^6+36u^3v^2w^6+3uv^4w^6\leq0$$ because $u\geq v\geq w$ and $v^4\geq uw^3$, which says that it's enough to prove $f(w^3)\geq0$ for a maximal value of $w^3$,

which happens for equality case of two variables.

Since $f(w^3)\geq0$ is homogeneous, we can assume $y=z=1$, which gives $$\frac{8}{\sqrt3}\sqrt{\frac{2(x+2)(x+1)^2(2x+1)^3}{3x^2+2x+1}}+\frac{2(x+2)(x-1)^2}{3(2x+1)}\geq10x^2+28x+10$$ or $$(x-1)^2(549x^6+2158x^5+2749x^4+1588x^3+551x^2+158x+23)\geq0.$$ Done!

Answer 2

Denote the difference $\text{LHS}-\text{RHS}$ by $S(A,B,C)$. We need to show that $S(A,B,C)\ge0$. We begin with a simple

Proposition. For $\alpha\in[0,\pi/2]$ we have the following inequality $$2\cos\frac{\pi-\alpha}4-\sqrt3\cos\frac{\pi-3\alpha}4<\sqrt3$$ Proof. $\frac{\pi-3\alpha}4\in[-\frac\pi8,\frac\pi4]$, thus $$2\cos\frac{\pi-\alpha}4-\sqrt3\cos\frac{\pi-3\alpha}4\le2-\sqrt3\cos\frac\pi4<\sqrt3.\quad\square$$

Now suppose $A\le\pi/2$ (such an angle always exists in a triangle; e.g. take $A$ to be the smallest one among $A,B,C$). Reformulate the desired inequality as $$\cos\frac A2+2\cos\frac{\pi-A}4\cos\frac{B-C}4\ge\sqrt3\cos^2\frac{B-C}4-\frac{\sqrt3}2+\sqrt3\cos\frac{\pi-3A}4\cos\frac{B-C}4$$ Let $t=\cos\frac{B-C}4\in(0,1]$. Consider the function $$f(x)=-\sqrt3t^2+(2\cos\frac{\pi-A}4-\sqrt3\cos\frac{\pi-3A}4)x$$ By the proposition at the beginning of this proof, The axis of symmetry of $f$ is $$x=\frac{2\cos\frac{\pi-A}4-\sqrt3\cos\frac{\pi-3A}4}{2\sqrt3}<\frac12$$ And since $-\sqrt3<0$, the parabola opens downwards. Consequently $f(t)\ge f(1)$. Furthermore, $f(t)>f(1)$ if $t\not=1$. In other words, $S(A,B,C)\ge S(A,\frac{B+C}2,\frac{B+C}2)$ with the equality if and only if $B=C$.

We're almost done. Observe that $S$ is continuous in $A, B, C$. We may extend the range of $A,B,C$ to $[0,\pi]^3\cap\{A+B+C=\pi\}$ without destroying the above argument. This is a compact set, on which $S$ attains its minimum $S_0$ at the point $(A_0, B_0, C_0)$. WLOG we may assume $A\le\pi/3$. If $B_0\not=C_0$, then $S_0>S(A_0,\frac{B_0+C_0}2,\frac{B_0+C_0}2)$, contradicting the minimality of $S_0$. Thus $B_0=C_0$. But $B_0+C_0=\pi-A_0\le\pi$, so $B_0\le\pi/2$. Applying the above argument once again, we see $A_0=C_0$. Now we conclude that $A_0=B_0=C_0=\pi/3$, hence $S_0=S(\pi/3,\pi/3,\pi/3)=0$. This shows that $S\ge S_0=0$.

Answer 3

We can make also the following thing.

We'll prove that for all triangle the following inequality is true. $$\sin\alpha+\sin\beta+\sin\gamma\geq\frac{\sqrt{3}}{2}(\cos(\alpha-\beta)+\cos(\alpha-\gamma)+\cos(\beta-\gamma)).$$ Indeed, let $R$ be a radius of circumcircle, $r$ be a radius of inscribed circle and $p$ be a semiperimeter of the triangle.

Hence, $$\sum_{cyc}\sin\alpha=\sum_{cyc}\frac{a}{2R}=\frac{p}{R}.$$ In another hand, easy to show that $$\sum_{cyc}\cos(\alpha-\beta)=\frac{p^2-2R^2+2Rr+r^2}{2R^2}.$$ Thus, we need to prove that $$\frac{p}{R}\geq\frac{\sqrt3(p^2-2R^2+2Rr+r^2)}{4R^2}$$ or $$\sqrt3p^2-4Rp-2\sqrt3R^2+2\sqrt3Rr+\sqrt3r^2\leq0$$ and since $$-2\sqrt3R^2+2\sqrt3Rr+\sqrt3r^2=-2\sqrt3R(R-r)+\sqrt3r^2\leq-2\sqrt3\cdot2r(2r-r)+\sqrt3r^2<0,$$ we need to prove that $$p\leq\frac{2R+\sqrt{10R^2-6Rr-3r^2}}{\sqrt3}.$$ Now, by Gerretsen inequality $p\leq\sqrt{4R^2+4Rr+3r^2}$.

Id est, it's enough to prove that $$\sqrt{4R^2+4Rr+3r^2}\leq\frac{2R+\sqrt{10R^2-6Rr-3r^2}}{\sqrt3}.$$ Let $R=2rx$.

Hence, $x\geq1$ and we need to prove that: $$\sqrt{16x^2+8x+3}\leq\frac{4x+\sqrt{40x^2-12x-3}}{\sqrt3}$$ or $$2x^2-9x-3+2x\sqrt{40x^2-12x-3}\geq0$$ or $$2x^2+x-3+2x\left(\sqrt{40x^2-12x-3}-5\right)\geq0$$ or $$(x-1)\left(2x+3+\frac{8x(10x+7)}{\sqrt{40x^2-12x-3}+5}\right)\geq0,$$ which is obvious.

Now, since the inequality $$\sin\alpha+\sin\beta+\sin\gamma\geq\frac{\sqrt{3}}{2}(\cos(\alpha-\beta)+\cos(\alpha-\gamma)+\cos(\beta-\gamma))$$ is true for all triangle, this inequality is true also for any acute-angled triangle

and after replacing in the last inequality

$\alpha$ on $90^{\circ}-\frac{\alpha}{2}$, $\beta$ on $90^{\circ}-\frac{\beta}{2}$ and $\gamma$ on $90^{\circ}-\frac{\gamma}{2}$ we'll get the starting inequality.

Done!