Proving that, for an acute $\triangle ABC$, $\sin A + \sin B+\sin C\gt \cos A+\cos B+\cos C$

Question

I need to prove or disprove that in any acute $\triangle ABC$, the following property holds: $$\sin A + \sin B + \sin C \gt \cos A + \cos B + \cos C$$

To begin, I proved a lemma:

Lemma. An acute triangle has at most one angle which is less than or equal to $\dfrac{\pi}{4}$.

Proof:

Let there be an acute angled $\Delta ABC$ with the angles $A$ & $B \le \frac{\pi}{4}$. Then

$$ A + B \le \frac{\pi}{2} \implies - (A + B) \ge -\frac{\pi}{2} \implies C = \pi - (A+B) \ge \frac{\pi}{2}$$ thus contradicting that the triangle is obtuse. Hence, by contradiction, the lemma is proved. $\square$

Further, I used the identity that $\sin x - \cos x = \sqrt{2}\sin (x - \frac{\pi}{4})$ to rewrite the inequality as

$$\sin \biggr(A - \frac{\pi}{4}\biggr) + \sin \biggr(B - \frac{\pi}{4}\biggr) + \sin \biggr(C - \frac{\pi}{4}\biggr) \gt 0$$

Without loss of generality, I assumed that $A \le \frac{\pi}{4}$.

If $A = \dfrac{\pi}{4}$, then the inequality follows, since both $B$ and $C$ are strictly greater than $\dfrac{\pi}{4}$.

How do I prove the inequality if $A \lt \dfrac{\pi}{4}$?

Any help or hint will be appreciated.

Answer 1

For a nice algebraic way to prove it, consider the inequality

$$\sin x\geq \frac{2x}{\pi}, x\in[0,\frac{\pi}{2}]$$

and apply for $A,B,C>\frac{\pi}{4}$:

$$\sin(A-\frac{\pi}{4})+\sin(B-\frac{\pi}{4})+\sin(C-\frac{\pi}{4})\geq\frac{2}{\pi}(A+B+C-\frac{3\pi}{4})=\frac{1}{2}>0$$

When we assume wlog $A<\frac{\pi}{4}$, we can instead assert that: $$\sin(A-\frac{\pi}{4})\geq A-\frac{\pi}{4}$$ and then

$$\sin(A-\frac{\pi}{4})+\sin(B-\frac{\pi}{4})+\sin(C-\frac{\pi}{4})\geq 1-\frac{\pi}{4}+(1-\frac{2}{\pi})A>1-\frac{\pi}{4}>0$$

This is not the sharpest bound. Actually one can show that the minimum value this expression is achieved for A=B=C and therefore

$$\require{cancel}\xcancel{\sin(A-\frac{\pi}{4})+\sin(B-\frac{\pi}{4})+\sin(C-\frac{\pi}{4})\geq 3\sin(\frac{\pi}{12})=\frac{3}{2}\sqrt{2-\sqrt{3}}}$$

EDIT: After a while, I finally noticed that, while the solution to the problem satisfactorily addresses the question asked, a mistake above has obscured a clear elementary solution to finding maxima and minima to this function. We want to show that whenever $A,B,C>0$ with $A+B+C=\pi$,

$$\frac{\sqrt{2}}{2}<\Delta=\sin(A-\frac{\pi}{4})+\sin(B-\frac{\pi}{4})+\sin(C-\frac{\pi}{4})\leq 3\sin\frac{\pi}{12}$$

In the figure-where the contours of $\Delta(A,B)$ are depicted- one can clearly see that, within the area of interest denoted by a red triangle, the function attains a maximum. The minima are attained on the red triangle itself. As of now I haven't found a satisfactory algebraic/elementary approach to obtain the minima and maxima in the entire triangle interior and boundary (but with calculus it is pretty easy to demonstrate).

To get the maximum in a part of the interior, assume that $A,B,C>\frac{\pi}{4}$. In this case we can apply Jensen's inequality for the concave function $f(x)=\sin(x)~,~ x\in (0,\pi)$ and we obtain $$\sin(A-\frac{\pi}{4})+\sin(B-\frac{\pi}{4})+\sin(C-\frac{\pi}{4})\leq 3\sin(\frac{A+B+C}{3}-\frac{\pi}{4})=3\sin\frac{\pi}{12}$$.

Answer 2

Hint: Use the formulas $$\cos(\alpha)+\cos(\beta)+\cos(\gamma)=\frac{r}{R}+1$$ $$2A=ab\sin(\gamma)=ac\sin(\beta)=ab\sin(\gamma)$$ $$A=sr=\frac{abc}{4R}$$ $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ Doing this you will get $$4A(a+b+c)>a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)$$

Answer 3

A purely "trig-bashing" algebraic way: We have \begin{align*} &\sin\left(B-\frac\pi4\right)+\sin\left(C-\frac\pi4\right)\\ &=2\sin\left(\frac{B+C}2-\frac\pi4\right)\cos\frac{B-C}2\\ &=2\sin\left(\frac\pi4-\frac{A}2\right)\cos\frac{B-C}2. \end{align*} So we want to prove $$ 2\cos\frac{B-C}2-1>0 $$ when $A<\pi/4$, $B,C$ acute. Note that we have $\lvert B-C\rvert$ is at most $A$ (only in the degenerate case when $B$ or $C$ is $\pi/2$), so $$ 2\cos\frac{B-C}2-1>2\cos\frac{A}2-1>2\cos\frac\pi8-1>0 $$ as desired.

Answer 4

Note that A,B,C are are positive acute angles, so both LHS and RHS are positive. Square both LHS and RHS, Then do LHS - RHS we get.

$\cos2A + \cos2B + \cos2C + 2(\cos(A+B)+\cos(A+C)+\cos(B+C))$ note that $2(\cos(A+B)+\cos(A+C)+\cos(B+C))$ is always -ve for acute triangle (all these angles fall in second quadrant).

Check $\cos2A + \cos2B + \cos2C$, If all are greater than 45 deg. then this becomes negative. Suppose $A= 45 -\theta$ (in deg.), writing

$\cos2(45 -\theta) + \cos2(45 +\theta + a) + \cos2(45 +\theta + b)$ where $0\leq \theta < 45$ , $0< a + \theta < 45$, $0< b + \theta < 45$

This part becomes

$\sin2\theta - \sin(2a+2\theta) - \sin(2b+2\theta)$, all these angles fall in first quadrant, so this is also negative. So

$LHS < RHS$

Answer 5

$\Sigma \sin A\ge\Sigma\cos A\iff\Sigma\sin\left (A-\frac{\pi}{4}\right)\ge 0.$ If all are greater than $\frac{\pi}{4}$, it is trivially true.

let $A<\frac {\pi}{4}<B\le C$

Now using identity $\Sigma \sin x-\sin\Sigma x=4\prod\sin\left(\frac{x+y}{2}\right)$, we get:

$\Sigma\sin\left (A-\frac{\pi}{4}\right)=\frac{1}{\sqrt 2}+4\prod\cos \left (A+\frac{\pi}{4}\right)\ge \frac{1}{\sqrt 2}>0$ Done!

Answer 6

Your problem is equivalent to the following: In any acute $ABC$ we have $AB+BC+CA>AH+BH+CH$ , where $H$ is the orthocenter of the triangle. But this inequality is true for all points from the tringle's interior.