2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

$\cos(\alpha)+\cos(\gamma)+\sin(\beta)=\sin(\frac{\pi}{2}-\alpha)+\sin(\frac{\pi}{2}-\gamma)+\sin(\beta)$
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 : $\cos(\alpha)+\cos(\gamma)+\sin(\beta) \leq 2 \sin(\frac{\pi}{4})+\sin(\beta)=1+\sqrt{2}\leq \frac{3\sqrt{3}}{2}$

$\endgroup$
  • $\begingroup$ 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 $x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ $\endgroup$ – Michael Rozenberg Oct 26 '16 at 12:46
0
$\begingroup$

Because if $\cos\frac{\beta}{2}=x$ by AM-GM we obtain: $$\cos\alpha+\cos\gamma+\sin\beta=2\sin\frac{\beta}{2}\cos\frac{\alpha-\gamma}{2}+\sin\beta\leq2\sin\frac{\beta}{2}+\sin\beta=$$ $$=2\sqrt{1-x^2}(1+x)=2\sqrt{(1-x)(1+x)^3}=2\sqrt{27(1-x)\left(\frac{1}{3}+\frac{x}{3}\right)^3}\leq$$ $$\leq2\sqrt{27\left(\frac{1-x+3\left(\frac{1}{3}+\frac{x}{3}\right)}{4}\right)^4}=\frac{3\sqrt3}{2}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.