MHB Review my solution: Trigonometry proof

AI Thread Summary
The discussion focuses on proving the inequality \( c \geq (a+b) \sin \left( \frac{C}{2} \right) \) for a triangle with sides \( a, b, c \) and angle \( C \) subtending side \( c \). The proof utilizes the law of cosines and properties of sine to establish that \( c^2 \) is greater than or equal to \( (a+b)^2 \sin^2 \left( \frac{C}{2} \right) \). An alternative proof is presented using the triangle's area and the relationship between the angle bisector and altitude, reinforcing the original conclusion. The participants agree on the validity of the proofs and emphasize the importance of multiple proofs in mathematics. Overall, the discussion highlights different approaches to a geometric inequality in trigonometry.
SweatingBear
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For a triangle with sides $$a$$, $$b$$, $$c$$ and angle $$C$$, where the angle $$C$$ subtends the side $$c$$, show that

$$c \geqslant (a+b) \sin \left( \frac C2 \right)$$
_______

Let us stipulate $$0^\circ < C < 180^\circ$$ and of course $$a,b,c > 0$$. Consequently, $$0^\circ < \frac C2 \leqslant 90^\circ \implies 0 < \sin^2 \left( \frac C2 \right) \leqslant 1$$ (we include $$90^\circ$$ for the angle $$\frac C2$$ in order to account for right triangles).

Law of cosines yield

$$c^2 = a^2 + b^2 - 2ab \cos ( C )$$

Using $$\cos (C) \equiv 1 - 2\sin^2 \left( \frac C2 \right)$$, we can write

$$c^2 = a^2 + b^2 - 2ab + 4ab\sin^2 \left( \frac C2 \right)$$

Now since $$0 < \sin^2 \left( \frac C2 \right) \leqslant 1$$, $$a^2 + b^2 - 2ab + 4ab\sin^2 \left( \frac C2 \right)$$ will either have to equal $$a^2 + b^2 - 2ab + 4ab$$ or be greater than it. Thus

$$c^2 \geqslant a^2 + b^2 - 2ab + 4ab = a^2 + 2ab + b^2 = (a+b)^2 $$

A similar argument can be made for $$(a+b)^2$$ versus $$(a+b)^2 \sin^2 \left( \frac C2 \right)$$: $$(a+b)^2$$ will either have to equal or be greater than the latter expression, due to the values $$\sin^2 \left( \frac C2 \right)$$ can assume. Therefore

$$c^2 \geqslant (a+b)^2 \geqslant (a+b)^2 \sin^2 \left( \frac C2 \right) $$

We can finally conclude

$$c \geqslant (a+b) \sin \left( \frac C2 \right)$$

Thoughts?
 
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I see nothing wrong with your proof. This is how I would prove it. Please refer to the following diagram:

View attachment 1085

I would postulate concerning the angle bisector $m$ of $\angle C$ and the altitude $h$, we must have:

(1) $$m\ge h$$

That is, the shortest distance between a point and a line is the perpendicular distance.

Now, the area $T$ of the triangle may be written in these ways:

$$T=\frac{1}{2}ch=\frac{1}{2}m(a+b)\sin\left(\frac{C}{2} \right)$$

Note: I have made use of the formulas (for a general triangle):

$$T=\frac{1}{2}bh$$

$$T=\frac{1}{2}ab\sin(C)$$

Thus, we have:

$$ch=m(a+b)\sin\left(\frac{C}{2} \right)$$

And from (1), we therefore conclude:

$$c\ge (a+b)\sin\left(\frac{C}{2} \right)$$
 

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MarkFL said:
I see nothing wrong with your proof. This is how I would prove it. Please refer to the following diagram:

View attachment 1085

I would postulate concerning the angle bisector $m$ of $\angle C$ and the altitude $h$, we must have:

(1) $$m\ge h$$

That is, the shortest distance between a point and a line is the perpendicular distance.

Now, the area $T$ of the triangle may be written in these ways:

$$T=\frac{1}{2}ch=\frac{1}{2}m(a+b)\sin\left(\frac{C}{2} \right)$$

Note: I have made use of the formulas (for a general triangle):

$$T=\frac{1}{2}bh$$

$$T=\frac{1}{2}ab\sin(C)$$

Thus, we have:

$$ch=m(a+b)\sin\left(\frac{C}{2} \right)$$

And from (1), we therefore conclude:

$$c\ge (a+b)\sin\left(\frac{C}{2} \right)$$

Thank for the feedback and the excellent alternative solution!
 
A good rule of thumb in mathematics I live by is this:

If something is true, you should be able to provide two proofs, and one of them should have a picture. This is a perfect example of what I mean.
 
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