Example - 9

Let \(x, y, z\) be positive real numbers such that \(\begin{align}x + y + z = 1\end{align}\). Determine the minimum value of

\[\begin{align}\frac{1}{x} + \frac{4}{y} + \frac{9}{z}\end{align}\]

Solution: An application of Cauchy-Schwarz inequality makes this is a one-step problem. Nevertheless, we present a proof which involves only the easier inequality \(\begin{align}{x^2} + {y^2} \ge \;2xy\end{align}\) for real numbers \(x\) and \(y,\) by setting first \(\begin{align}x = \tan b\;{\rm{and}}\;y = 2\tan b\end{align}\) and second \(\begin{align}x = \tan a\;{\rm{and}}\;y = \cot a\end{align}\).

Clearly, \(z\) is a real number in the interval [0, 1]. Hence there is an angle a such that \(\begin{align}z = {\sin ^2}a.\end{align}\) Then \(\begin{align}x + y = 1 - {\sin ^2}a = {\cos ^2}a,\;{\rm{or}}\;\frac{x}{{{{\cos }^2}a}} + \frac{y}{{{{\cos }^2}a}} = 1\end{align}\). For an angle \(b,\) we have \(\begin{align}{\cos ^2}b + {\sin ^2}b = 1\end{align}\). Hence, we can set \(\begin{align}x = {\cos ^2}a\;{\cos ^2}b,\;y = {\cos ^2}a\;{\sin ^2}b\end{align}\)  for some angle \(b.\) 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

\[\begin{align}    P & = 14 + 5{\tan ^2}a + 9{\cot ^2}a + ({\tan ^2}b + 4{\cot ^2}b)(1 + {\tan ^2}a)\\    & > 14 + 5{\tan ^2}a + 9{\cot ^2}a + 2\tan b \cdot 2\cot b(1 + {\tan ^2}a)\\    & = 18 + 9({\tan ^2}a + {\cot ^2}a) \ge 18 + 9 \cdot 2\tan a\cot a = 36   \end{align}\]

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

 

Example - 10

Let \(A_1 \)be the center of the square inscribed in acute triangle \(ABC\) with two vertices of the square on side \(BC\) (shown in figure). Thus one of the two remaining vertices of the square lies on side \(AB\) and the other on segment \(AC.\) Points \(B_1\) and \(C_1\) are defined in a similar way for inscribed squares with two vertices on sides \(AC\) and \(AB,\) respectively. Prove that lines \(AA_1, BB_1, CC_1 \)are concurrent.

Solution:

Let line \(AA_1\) and segment \(BC\) intersect at \(A_2.\) We define \(B_2\) and \(C_2\) analogously. By Ceva’s theorem, it suffices to show that

\[\begin{align}\frac{{\sin \angle BA{A_2}}}{{\sin \angle {A_2}AC}} \cdot \frac{{\sin \angle CB{B_2}}}{{\sin \angle {B_2}BA}} \cdot \frac{{\sin \angle AC{C_2}}}{{\sin \angle {C_2}CB}} = 1.\end{align}\]

Let the vertices of the square be \(DETS,\) labled as shown in figure. Applying the law of sines to triangles \(\begin{align}AS{A_1}\end{align}\) and \(\begin{align}AT{A_1}\end{align}\) gives

\[\begin{align}\frac{{|A{A_1}|}}{{|S{A_1}|}} = \frac{{\sin \angle AS{A_1}}}{{\sin \angle SA{A_1}}} = \frac{{\sin \angle AS{A_1}}}{{\sin \angle BA{A_2}}}\;{\rm{and}}\;\frac{{|T{A_1}|}}{{|A{A_1}|}} = \frac{{\sin \angle {A_1}AT}}{{\sin \angle AT{A_1}}} = \frac{{\sin \angle {A_2}AC}}{{\sin \angle AT{A_1}}}\end{align}\]

Because \(\begin{align}|{{A}_{1}}S|=|{{A}_{1}}T|\ \text{and}\ \angle AS{{A}_{1}}=B+45{}^\text{o}\ \text{and}\ \angle AT{{A}_{1}}=C+45{}^\text{o}\end{align}\), multiplying the above identities yields

\[\begin{align}1=\frac{|A{{A}_{1}}|}{|S{{A}_{1}}|}\cdot \frac{|T{{A}_{1}}|}{|A{{A}_{1}}|}-\frac{\sin \angle AS{{A}_{1}}}{\sin \angle BA{{A}_{2}}}\cdot \frac{\sin \angle {{A}_{2}}AC}{\sin \angle AT{{A}_{1}}}\end{align}\]

implying that

\[\begin{align}\frac{\sin \angle BA{{A}_{2}}}{\sin \angle {{A}_{2}}AC}=\frac{\sin \angle AS{{A}_{2}}}{\sin \angle AT{{A}_{1}}}=\frac{\sin (B+45{}^\text{o})}{\sin (C+45{}^\text{o})}\end{align}\]

In exactly the same way, we can show that

\[\begin{align}\frac{\sin \angle CB{{B}_{2}}}{\sin \angle {{B}_{2}}BA}=\frac{\sin (C+45{}^\text{o})}{\sin (A+45{}^\text{o})}\ \ \text{and}\ \ \frac{\sin \angle AC{{C}_{2}}}{\sin \angle {{C}_{2}}CB}=\frac{\sin (A+45{}^\text{o})}{\sin (B+45{}^\text{o})}\end{align}\]

Multiplying the last three identities establishes the desired result.

 

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