1. ## Inequalities(3):

For any acute angle $\displaystyle \Delta ABC$ , prove that

$\displaystyle \frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}\le\frac{9\sqrt3}{2\pi}$

2. Using Jensen's inequality for the function $\displaystyle \frac{sin x}{x}$,

$\displaystyle \frac{\frac{sin A}{A}+\frac {sin B}{B}+\frac{sin C}{C}}{3} \leq \frac{sin \frac{A+B+C}{3}}{\frac{A+B+C}{3}}$

$\displaystyle \frac{\frac{sin A}{A}+\frac {sin B}{B}+\frac{sin C}{C}}{3} \leq \frac{sin \frac{\pi}{3}}{\frac{\pi}{3}}$

$\displaystyle \frac{\frac{sin A}{A}+\frac {sin B}{B}+\frac{sin C}{C}}{3} \leq \frac{\frac{\sqrt{3}}{2}}{\frac{\pi}{3}}$

$\displaystyle \frac{sin A}{A}+\frac {sin B}{B}+\frac{sin C}{C} \leq 3 \frac{\frac{\sqrt{3}}{2}}{\frac{\pi}{3}}$

$\displaystyle \frac{sin A}{A}+\frac {sin B}{B}+\frac{sin C}{C} \leq \frac{9\sqrt3}{2\pi}$

3. Please check the question....I think it is incorrect.

.........my mistake ::::Confused A and a

How to delete the post???

4. Let $\displaystyle f(x)=\frac{\sin x}{x}$

$\displaystyle f'(x)=\frac{x\cos x-\sin x}{x^2}=\frac{\cos x(x-\tan x)}{x^2}<0$,since $\displaystyle x<\tan x$ for $\displaystyle 0<x<\frac{\pi}{2}$

$\displaystyle f''(x)=\frac{-x^2\sin x-2x\cos x+2\sin x}{x^3}$

Let $\displaystyle g(x)=-x^2\sin x-2x\cos x+2\sin x$

$\displaystyle g'(x)=-x^2\cos x<0$ for $\displaystyle 0<x<\frac{\pi}{2}$

For $\displaystyle x>0$,we have $\displaystyle g(x)<g(0)$ and thus $\displaystyle g(x)<0$

Therefore, $\displaystyle f''(x)<0$ for all $\displaystyle x\in \left(0,\frac{\pi}{2}\right)$.

Thus the function y=f(x) is concave downwards for all $\displaystyle x\in \left(0,\frac{\pi}{2}\right)$.

$\displaystyle \therefore f\left(\frac{A+B+C}{3}\right)\geq \frac{f(A)+f(B)+f(C)}{3}$

$\displaystyle \frac{\sin\left(\frac{A+B+C}{3}\right)}{\frac{A+B+ C}{3}}\geq \frac{\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}}{3}$

$\displaystyle \frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}\leq \frac{9\sqrt{3}}{2\pi}$

5. Originally Posted by alexmahone
Using Jensen's inequality for the function $\displaystyle \frac{sin x}{x}$,

$\displaystyle \frac{\frac{sin A}{A}+\frac {sin B}{B}+\frac{sin C}{C}}{3} \leq \frac{sin \frac{A+B+C}{3}}{\frac{A+B+C}{3}}$
$\displaystyle \frac{\frac{sin A}{A}+\frac {sin B}{B}+\frac{sin C}{C}}{3} \leq \frac{sin \frac{\pi}{3}}{\frac{\pi}{3}}$
$\displaystyle \frac{\frac{sin A}{A}+\frac {sin B}{B}+\frac{sin C}{C}}{3} \leq \frac{\frac{\sqrt{3}}{2}}{\frac{\pi}{3}}$
You first of all need to prove that $\displaystyle f''(x)<0, \forall x\in\left(0,\frac{\pi}{2}\right)$ to use the Jensen's inequality

6. Originally Posted by pankaj
You first of all need to prove that $\displaystyle f''(x)<0, \forall x\in\left(0,\frac{\pi}{2}\right)$ to use the Jensen's inequality
I know that, lol.

7. Originally Posted by pankaj

Thus the function y=f(x) is concave downwards for all $\displaystyle x\in \left(0,\frac{\pi}{2}\right)$.

$\displaystyle \therefore f\left(\frac{A+B+C}{3}\right)\geq \frac{f(A)+f(B)+f(C)}{3}$

How inequality follows from the condition?

8. Originally Posted by malaygoel
How inequality follows from the condition?
Consider the three points on the curve $\displaystyle (A,f(A))$ $\displaystyle (B,f(B))$ and $\displaystyle (C,f(C))$. When joined they form a triangle which obviously lies below the curve $\displaystyle y=f(x)$.Now centroid of this triangle is inside the triangle and thus centroid is also below the curve.

Co-ordinates of centroid=$\displaystyle \left(\frac{A+B+C}{3},\frac{f(A)+f(B)+f(C)}{3}\rig ht)$.

The corresponding point on the curve with the same $\displaystyle x$-coordinate will be $\displaystyle \left(\frac{A+B+C}{3},f(\frac{A+B+C}{3})\right)$

Since this point is above the centroid .Therefore, $\displaystyle f(\frac{A+B+C}{3})>\frac{f(A)+f(B)+f(C)}{3}.$

Hey,you can try plotting $\displaystyle y=\ln x$ and draw a triangle and check for yourself

9. Originally Posted by great_math
For any acute angle $\displaystyle \Delta ABC$ , prove that

$\displaystyle \frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}\le\frac{9\sqrt3}{2\pi}$
Does the information that triangle is acute is of any help here? The proofs seem to be valid for obtuse triangles as well.