Trigonometric substitution in inequalities


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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