MHB Trigonometric inequality challenge

[Albert1]

Acute triangle ABC
Prove :Sin A +Sin B +Sin C>Cos A + Cos B + Cos C

 

[MarkFL]

My solution:

Spoiler

Consider the objective function:

$$F(A,B,C)=\sin(A)+\sin(B)+\sin(C)-\cos(A)-\cos(B)-\cos(C)$$

Subject to the constraint:

$$g(A,B,C)=A+B+C-\pi=0$$

Because of cyclic symmetry, we know the critical point occurs for:

$$A=B=C=\frac{\pi}{3}$$

We find:

$$f\left(\frac{\pi}{3},\frac{\pi}{3},\frac{\pi}{3}\right)=\frac{3}{2}(\sqrt{3}-1)\approx1.098$$

and:

$$f\left(\frac{\pi}{3},\frac{5\pi}{12},\frac{\pi}{4}\right)=\frac{1}{2}(\sqrt{3}+\sqrt{2}-1)\approx1.073$$

Now, if we let one of the angles be a right angle, then the other two will be complementary, and we will have:

$$f=1$$

Therefore, for an acute triangle, we have:

$$1<f\le\frac{3}{2}(\sqrt{3}-1)$$

And from this we conclude the given inequality is true.

 

[Albert1]

Acute triangle ABC
Prove :Sin A +Sin B +Sin C>Cos A + Cos B + Cos C

hint:

Spoiler

$\angle A+\angle B>90^o---(1)$
$\angle B+\angle C>90^o---(2)$
$\angle C+\angle A>90^o---(3)$
$why?$

 

[lfdahl]

Suggested solution:

Spoiler

\[A+B+C = 180^{\circ} \: \: \: \wedge \: \: \: C < 90^{\circ}\Rightarrow A+B > 90^{\circ}\]

Since all three angles (in an acute triangle) are less than $90^{\circ}$, we also have:\[A+C > 90^{\circ},\: \: \: B+C > 90^{\circ}\]Using the fact, that $ \sin X $ is a monotonically increasing function for $0^{\circ} \le X \le 90^{\circ}$, we get:

\[A + B > 90^{\circ} \Rightarrow \sin A>\sin(90^{\circ}-B)= \cos B\]

Similarly:

\[B+C > 90^{\circ} \Rightarrow \sin B>\sin(90^{\circ}-C)= \cos C\]

\[A+C > 90^{\circ} \Rightarrow \sin C>\sin(90^{\circ}-A)= \cos A\]

Adding the three inequalities yields the result:

\[\sin A+\sin B+\sin C > \cos A+\cos B+\cos C\]