Κυριακή, 11 Ιουνίου 2017

Trigonometric substitution in inequalities | My Two Cents

Trigonometric substitution in inequalities | My Two Cents




Trigonometric substitution in inequalities




 
 
 
 
 
 
5 Votes

Trigonometry
and inequalities are my 2 favourite topics in MO. It gets better when
both of them are involved together. Before we begin, let us attempt some
simple warm up questions that may even appear in our Sec 4 mathematics
syllabus:
For triangle ABC, prove that
1) \tan{A}+\tan{B}+\tan{C}=\tan{A}\tan{B}\tan{C}
2) \tan{\dfrac{A}{2}}\tan{\dfrac{B}{2}}+\tan{\dfrac{B}{2}}\tan{\dfrac{C}{2}}+\tan{\dfrac{C}{2}}\tan{\dfrac{A}{2}}=1
3) \sin^2 {\dfrac{A}{2}}+\sin^2 {\dfrac{B}{2}}+\sin^2 {\dfrac{C}{2}}+2\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}=1
One can get the above results upon
application of compound angle formulas. However, don’t expect to see
these sort of the problems in SMO since these identities are widely
publicised in Olympiad materials already. Instead, these identities will
be applied in other fields of MO, most notably in solving inequalities.
So for example, if the constraint x+y+z=xyz is given, one can substitute x=\tan{A}, y=\tan{B}, z=\tan{C} such that A+B+C=\pi.
After which, one can use trigonometric identities to simplify the
equation or apply Jensen’s inequality on the trigonometric functions.
Let us see an application of the strategy in SMO(O)2011 2nd Round P3:
Suppose x,y,z>0 and \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}<\dfrac{1}{xyz}, prove that
\dfrac{2x}{\sqrt{1+x^2}}+\dfrac{2y}{\sqrt{1+y^2}}+\dfrac{2z}{\sqrt{1+z^2}}<3
Here’s my solution to the problem during
the competition itself. If you have tried using trigonometric
substitution before, you will realise that this question is SCREAMING
for you to substitute the variables using tangent functions (possible to
use cotangent function too). So let’s do that. From the condition, we
can rearrange the constraint into xy+yz+zx<1. Now let’s introduce our tangent functions, substitute and obtaining \tan{\dfrac{A}{2}}\tan{\dfrac{B}{2}}+\tan{\dfrac{B}{2}}\tan{\dfrac{C}{2}}+\tan{\dfrac{C}{2}}\tan{\dfrac{A}{2}}<1, which suggests that A+B+C<\pi.
The inequality that we want to prove can will be simplified into an elegant expression after substitution. We have
\dfrac{2x}{\sqrt{1+x^2}} = 2\sin{\dfrac{A}{2}}
So the inequality transforms into 2\sin{\dfrac{A}{2}}+2\sin{\dfrac{B}{2}}+2\sin{\dfrac{C}{2}}<3, which is easily proven using Jensen’s inequality since the sine function is concave for \theta<\pi.

In addition to this example,
trigonometric substitution may be useful if a bounded constraint is
given. For example, if it is stated in the question that x_i \in [-1,1], consider substituting x using a sine or cosine function.
There are also situations where you do
not even need any constraints at all to use trigonometric substitution!
Let us look at the following problem:
Prove that (ab+bc+ca-1)^2 \le (a^2+1)(b^2+1)(c^2+1) for real numbers a, b, c.
It doesn’t look feasible to apply mean
inequalities here because the variables take negative values as well. On
the other hand, the expression a^2+1 looks familiar. Sometimes, it is helpful to use trigonometric substitution in inequalities that contain the term a^2+1 as it can be used to simplify trigonometric expressions. Using tangent function, we obtain
(ab+bc+ca-1)^2 \le \sec^2 x\sec^2 y \sec^2 z
\Rightarrow (ab+bc+ca-1)^2\cos^2 x\cos^2 y\cos^2 z \le 1
By using compound angle formulas, we are able to simplify the left hand side of the equation into \cos^2(x+y+z) (try it!), which is obviously smaller than 1.

I shall conclude this post with one of my favourite MO problems:
Prove that \dfrac{x_1}{1+x_1^2}+\dfrac{x_2}{1+x_1^2+x_2^2}+\cdots+\dfrac{x^n}{1+x_1^2+x_2^2+\cdots+x_n^2}<\sqrt{n} where x_i are real numbers.
This problem can be solved beautifully
using using trigonometric substitution. Will give reward to HC students
who can find the correct substitution and solve the inequality 🙂 (Well it can be solved using other methods too but the trigo method is really nice)
Cheers,
ksj

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