2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 2
    $\begingroup$ @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
1
$\begingroup$

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 $$4A(a+b+c)>a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)$$

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ You have to plug in all these terms in the given inequality $\endgroup$ – Dr. Sonnhard Graubner May 25 at 5:55
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$LHS < RHS$

$\endgroup$

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.