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I know it's easy to prove with the help of linear inequalities, but this time I want to prove it with the help of trigonometry. Is it possible? If yes, then how?

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    $\begingroup$ What are linear inequalities? $\endgroup$ – velut luna Jul 31 '16 at 10:20
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    $\begingroup$ I wonder how many of the answers (including mine) implicitly use the fact that any two sides of the triangle is greater than the third one in the invoked theorems. $\endgroup$ – user261263 Jul 31 '16 at 16:03
  • $\begingroup$ This specific inequality has a well-known name, the Triangle inequality. Quite important as a easy to understand special case of the Cauchy–Schwarz inequality. $\endgroup$ – MSalters Aug 1 '16 at 14:48
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Use cosine law

$a^2=b^2 + c^2 -2bc \cos(A) \lt b^2 + c^2 + 2bc=(b+c)^2$ because $-1 \lt \cos(A) \lt 1$

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If they weren't, then the two sides would not manage to reach each other on paper. Draw a line and two circles with mid points at the end points of the line, if the sum of radiuses are not greater than the length of the line the circles will never intersect and you can not build a triangle.

Now behold my godly paint skills. If $c+b \lt a$ then no matter how we choose angles rotating around the end points of $a$ will we be able to make $b$ and $c$ ends meet to form a triangle. enter image description here

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    $\begingroup$ This proof requires almost no mental effort at all to understand - I wish more proofs were like this $\endgroup$ – Mark Aug 1 '16 at 0:45
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    $\begingroup$ @Mark: because it's an intuitive proof. For example it doesn't work on the surface of a sphere (in which case the circles could intersect "around the back"), and it doesn't explicitly state what properties of the Euclidean plane it's using to distinguish it from a sphere. $\endgroup$ – Steve Jessop Aug 1 '16 at 8:43
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    $\begingroup$ Very nice and intuitive proof, but not much trigonometry inside, like OP requested. $\endgroup$ – user261263 Aug 1 '16 at 18:59
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    $\begingroup$ Wow. I honestly admire your mspaint skills! $\endgroup$ – dbanet Aug 2 '16 at 1:28
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A classic application of the incircle: enter image description here

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    $\begingroup$ Very elegant, but it would be nice if you spelt out the proof. $\endgroup$ – Jack Aidley Aug 1 '16 at 12:24
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    $\begingroup$ @JackAidley: You mean x+2*z+y > x+yor more simply 2*z >0 ? $\endgroup$ – MSalters Aug 1 '16 at 14:44
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    $\begingroup$ Yes, that's the essence of it. But a few words to that effect would make it immediately obvious to the reader how the graph elegantly translates to a proof instead of taking a few moments to deduce. $\endgroup$ – Jack Aidley Aug 1 '16 at 15:03
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    $\begingroup$ What's the proof for why those two incircle created segments should be equal? Pardon my ignorance if that is obvious. $\endgroup$ – curious_cat Aug 1 '16 at 15:54
  • $\begingroup$ @curious_cat Two tangents to a circle from a point are equal: using Pythagorean Theorem, their squares are equal to (distant to center$)^2 - ($radius$)^2$. $\endgroup$ – Quang Hoang Aug 1 '16 at 16:01
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enter image description here

$c = b \cos A + a \cos B < a+b $

Since $\cos A < 1$ and $\cos B < 1$ and they cannot equal 1 simultaneously.

If $A$ or $B$ is obtuse, WLOG, let $A$ obtuse,$c = a \cos B + b \cos A < a+b $

Because $ \cos A < 0 < 1$ and $\cos B < 1$

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    $\begingroup$ For those confused: WLOG means Without loss of generality $\endgroup$ – user1717828 Jul 31 '16 at 15:10
  • $\begingroup$ How do we know that the foot of the perpendicular is between A and B? $\endgroup$ – user52817 Jul 31 '16 at 21:48
  • $\begingroup$ @user52817 easy. Since in my picture A is a ancute and B is an acute, then the only way to gule two triangle with right angle ,in order to form a triangle, is to use the common edge. So the foot of perpendicular is between A and B. $\endgroup$ – Zack Ni Jul 31 '16 at 23:03
  • $\begingroup$ Why can we assume the angle at A is acute? $\endgroup$ – user52817 Jul 31 '16 at 23:29
  • $\begingroup$ @user52817 no I am not assume, there is two case: A or B is an obtuse angle or right, and A and B are acute angles. $\endgroup$ – Zack Ni Jul 31 '16 at 23:31
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Perhaps not as directly-trigonometric as you want (@Eugen gave the answer I would've given in that regard), but ...

Heron's Formula. If $T$ is the area of the triangle with side-lengths $a$, $b$, $c$, then $$\frac{16\;T^2}{a+b+c} = (-a+b+c)(a-b+c)(a+b-c) \tag{$\star$}$$

Note that each factor on the right-hand side corresponds to an aspect of the Triangle Inequality. For non-degenerate triangles ($T>0$), the left-hand side is strictly positive, which implies that the number of negative factors on the right must be even; but, one readily determines that this number of factors cannot be two, so it must be zero, which is to say: all three Triangle Inequalities must hold. (I'll leave it to the reader to consider the degenerate case ($T=0$).)

Another way to think about this is:

Three lengths form a triangle if and only if Heron calculates a real area ($T$) from them. That is, Heron's formula not only computes a triangle's area, it determines a potential triangle's viability.

FYI: Menger's Theorem characterizing when six lengths form a tetrahedron works similarly: (1) Heron must calculate four real face areas, and (2) the Cayley-Menger determinant must calculate a real volume.

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Not an answer "using trigonometry" but one folks should know.

Direct from Euclid's Elements (Book I, Prop 20)

http://aleph0.clarku.edu/~djoyce/elements/bookI/propI20.html

enter image description here

(picture from wikipedia: https://en.wikipedia.org/wiki/Triangle_inequality#Euclidean_geometry)

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Maybe not exactly using trigonometry, but just using the definition of a line segment (shortest path between two given points) implies it directly. In particular for the triangle ABC, the path |AB| is shorter than any path between A and B including |AB|+|BC|.

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    $\begingroup$ So it does not use trigonometry "exactly" or at all. It seems you merely telling the OP something that was known at the time of the Question. See other Answers for ways that trigonometry might be used in an essential way. $\endgroup$ – hardmath Jul 31 '16 at 15:20
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    $\begingroup$ @hardmath Well, that is true my suggestion doesn't involve trigonometry. But any solution that contains a mapping between the edges of the triangle and a real number (length), requires the definition of the line segment which almost immediately implies the inequality. Since the definition of sine or cosine requires division of the lengths, I believe using them in an essential way is not possible. The essential proofs are the ones based on drawings such as Euclid's as proposed above. And thanks for your comment, I will be more careful to address the questions more carefully. $\endgroup$ – anil k. Jul 31 '16 at 15:55

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