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

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.

## 4 Answers

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

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

• But you are assuming $A,B,C$ all $\geq \pi/4$? – user10354138 May 25 at 5:47
• That inequality is a curious result; thanks for showing it to me – Aaratrick May 25 at 5:48
• I'll edit to cover all the cases. My bad – DinosaurEgg May 25 at 5:55
• @Aaratrick: It seems that the first inequality in this answer cannot be used when $x=A-\frac{\pi}{4}$ with $A\lt\frac{\pi}{4}$. – mathlove May 25 at 5:55
• @DinosaurEgg I believe that the inequality is reversed, i.e, $\sin x \le \frac{2x}{\pi}, x \in [-\frac{\pi}{2}, 0]. I got this by drawing the graph. – Aaratrick May 25 at 6:14 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 $$4A(a+b+c)>a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)$$ • I think in the first identity, the last term must be$\cos (\gamma)\$. Also could you please elaborate on this a bit more; I don't understand how to get the inequality from this – Aaratrick May 25 at 5:33
• You have to plug in all these terms in the given inequality – Dr. Sonnhard Graubner May 25 at 5:55

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.

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

$$LHS < RHS$$