# Prove the following Inequality.

Given that $\alpha,\beta$ and $\gamma\in (0,\pi)$ and $\alpha+\gamma+\beta=\pi$, show that $$\cos\alpha+\cos\gamma+\sin\beta\leq\frac{3\sqrt3}{2}.$$ Now I am aware of the following Inequality: $\cos\alpha+\cos\gamma+\cos\beta\leq\frac{3}{2}$, but notice that the term $\cos\beta$ has been replaced with $\sin\beta$. I also tried substituting $\sin\beta=\sin(\pi-(\alpha+\gamma))=\sin(\alpha+\gamma)$ and expanding the resulting expression but I was unable to deduce anything meaningful.

$\cos(\alpha)+\cos(\gamma)+\sin(\beta)=\sin(\frac{\pi}{2}-\alpha)+\sin(\frac{\pi}{2}-\gamma)+\sin(\beta)$
1) Suppose $\beta \leq \frac{\pi}{2}$
Using Jensen inequality we get: $\frac{\sin(\frac{\pi}{2}-\alpha)+\sin(\frac{\pi}{2}-\gamma)+\sin(\beta)}{3} \leq \sin(\frac{\frac{\pi}{2}-\alpha+\frac{\pi}{2}-\gamma+\beta}{3})=\sin(\frac{2\beta}{3})$
Since $\frac{2\beta}{3}\leq \frac{\pi}{3}$, we get : $\cos(\alpha)+\cos(\gamma)+\sin(\beta) \leq 3 \sin(\frac{\pi}{3})=\frac{3\sqrt{3}}{2}$
1) Suppose $\beta \geq \frac{\pi}{2}$
Using Jensen inequality we get: $\frac{\sin(\frac{\pi}{2}-\alpha)+\sin(\frac{\pi}{2}-\gamma)}{2} \leq \sin(\frac{\frac{\pi}{2}-\alpha+\frac{\pi}{2}-\gamma}{2})=\sin(\frac{\pi-\beta}{2})$
Since $\frac{\pi-\beta}{2} \leq \frac{\pi}{4}$, we get : $\cos(\alpha)+\cos(\gamma)+\sin(\beta) \leq 2 \sin(\frac{\pi}{4})+\sin(\beta)=1+\sqrt{2}\leq \frac{3\sqrt{3}}{2}$
• Your solution is total wrong! Let $f(x)=\cos x$. Hence, $f$ is not concave function and $f$ is not convex function. You can not use Jensen for $\beta\leq\frac{\pi}{2}$. Also if you want, $g(x)=\sin x$ is not concave function for $x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ – Michael Rozenberg Oct 26 '16 at 12:46
Because if $\cos\frac{\beta}{2}=x$ by AM-GM we obtain: $$\cos\alpha+\cos\gamma+\sin\beta=2\sin\frac{\beta}{2}\cos\frac{\alpha-\gamma}{2}+\sin\beta\leq2\sin\frac{\beta}{2}+\sin\beta=$$ $$=2\sqrt{1-x^2}(1+x)=2\sqrt{(1-x)(1+x)^3}=2\sqrt{27(1-x)\left(\frac{1}{3}+\frac{x}{3}\right)^3}\leq$$ $$\leq2\sqrt{27\left(\frac{1-x+3\left(\frac{1}{3}+\frac{x}{3}\right)}{4}\right)^4}=\frac{3\sqrt3}{2}$$