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 $\dfrac{\pi}{4}$.


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.


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


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

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

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    $\begingroup$ But you are assuming $A,B,C$ all $\geq \pi/4$? $\endgroup$ – user10354138 May 25 at 5:47
  • $\begingroup$ That inequality is a curious result; thanks for showing it to me $\endgroup$ – Aaratrick May 25 at 5:48
  • $\begingroup$ I'll edit to cover all the cases. My bad $\endgroup$ – DinosaurEgg May 25 at 5:55
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    $\begingroup$ @Aaratrick: It seems that the first inequality in this answer cannot be used when $x=A-\frac{\pi}{4}$ with $A\lt\frac{\pi}{4}$. $\endgroup$ – mathlove May 25 at 5:55
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    $\begingroup$ @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. $\endgroup$ – 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 $$4A(a+b+c)>a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)$$

  • $\begingroup$ 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 $\endgroup$ – Aaratrick May 25 at 5:33
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    $\begingroup$ You have to plug in all these terms in the given inequality $\endgroup$ – 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



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