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

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