A Trigonometry problem for IITJEE Advanced Mathematics or Mathematics Olympiad

Ref: Titu Andreescu and Zumin Feng.

Problem:

Let x, y, z be positive real numbers such that x+y+z=1. Determine the minimum value of

\frac{1}{x}+\frac{4}{y}+\frac{9}{z}

Solution:

An application of CauchySchwarz inequality makes this as a one step problem. Nevertheless, we present a proof which involves only the easier inequality x^{2}+y^{2} \geq 2xy for real numbers x and y by setting first x=\tan {b} and y=2\tan{b} and second x=\tan{a} and y=\cot{a}.

Clearly, z is a real number in the interval [0,1]. Hence, there is an angle a such that z=\sin^{2} {a}. Then, x+y=1-\sin^{2}{a}=\cos^{2}{a}, or \frac{x}{\cos^{2}{a}}+\frac{y}{\cos^{2}{a}}=1. For an angle b, we have \cos^{2}{b}+\sin^{2}{b}=1. Hence, we can set x=\cos^{2}{a}\cos^{2}{b}, and y=\cos^{2}{a}\sin^{2}{b} for some b, it suffices to find the minimum value of

P = \sec^{2}{a} \sec^{2}{b}+4\sec^{2}{a}\csc^{2}{b}+9\csc^{2}{a}

or

P = (\tan^{2}{a}+1)(\tan^{2}{b}+1)+4(\tan^{2}{a}+1)(\cot^{2}{b}+1)+9(\cot^{2}{a}+1)

Expanding the right hand side gives

P = 14 + 5\tan^{2}{a}+9\cot^{2}{a}+(\tan^{2}{b}+4\cot^{b})(1+\tan^{2}{a})

\geq 14 + 5\tan^{2}{a}+9\cot^{2}{a}+2\tan{b}.2\cot{b}(1+\tan^{2}{a})

= 18 + 9(\tan^{2}{a}+\cot^{2}{a}) \geq 18 + 9.2\tan{a}\cot{a}=36

Equality holds when \tan{a}=\cot{a} and \tan{b}=2\cot{b}, which implies that \cos^{2}{a}=\sin^{2}{a} and 2\cos^{2}{b}=\sin^{b}. Because \sin^{2}{\theta}+\cos^{2}{\theta}=1, equality holds when \cos^{2}{a}=\frac{1}{2} and \cos^{2}{b}=\frac{1}{3}, that is, x=\frac{1}{6}, y=\frac{1}{3}, z=\frac{1}{2}.

More later.

Nalin Pithwa

 

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: