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Test of Mathematics Solution Subjective 83 - Two numbers adding up to 1

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 83 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Problem

If a and b are positive real numbers such that a + b = 1, prove that $ \displaystyle \left( a + \frac{1}{a} \right)^2 $ + \( \left( b + \frac{1}{b} \right)^2 \ge \frac{25}{2} \)


Solution:

We replace 1 by a+b and expand the squares.

$ \displaystyle \left( a + \frac{a+b}{a} \right)^2 + \left(b + \frac{a+b}{b} \right)^2 $
$ \displaystyle \left( a + \frac{b}{a} + 1 \right)^2 + \left(b + \frac{a}{b} + 1 \right)^2 $
$ \displaystyle \left( a^2 + \frac{b^2}{a^2} + 1 + 2a + 2b + 2\frac{b}{a} \right) + \left(b^2 + \frac{b^2}{a^2} + 1 + 2b + 2a + 2 \frac{a}{b} \right) $

Now we use A.M. - G.M. inequality to have

$ \displaystyle { \frac {\frac{b^2}{a^2} + \frac{a^2}{b^2}}{2}\ge \sqrt {\frac{b^2}{a^2} \times \frac{a^2}{b^2}} = 1 } $
$ \displaystyle { \frac {\frac{b}{a} + \frac{a}{b}}{2}\ge \sqrt {\frac{b}{a} \times \frac{a}{b}} = 1 } $

Also we apply the square mean inequality to have

$ \displaystyle {\frac{a^2 + b^2}{2} \ge \left(\frac{a+b}{2} \right)^2} $ Since a+b = 1, we have $ a^2 + b^2 \ge \frac{1}{2} $

Combining all of them we have $ \displaystyle {\left( a + \frac{1}{a} \right)^2 + \left(b + \frac{1}{b} \right)^2 \ge 12 \frac{1}{2} = \frac{25}{2}} $


Discussion

Some theoretical background.

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 83 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Problem

If a and b are positive real numbers such that a + b = 1, prove that $ \displaystyle \left( a + \frac{1}{a} \right)^2 $ + \( \left( b + \frac{1}{b} \right)^2 \ge \frac{25}{2} \)


Solution:

We replace 1 by a+b and expand the squares.

$ \displaystyle \left( a + \frac{a+b}{a} \right)^2 + \left(b + \frac{a+b}{b} \right)^2 $
$ \displaystyle \left( a + \frac{b}{a} + 1 \right)^2 + \left(b + \frac{a}{b} + 1 \right)^2 $
$ \displaystyle \left( a^2 + \frac{b^2}{a^2} + 1 + 2a + 2b + 2\frac{b}{a} \right) + \left(b^2 + \frac{b^2}{a^2} + 1 + 2b + 2a + 2 \frac{a}{b} \right) $

Now we use A.M. - G.M. inequality to have

$ \displaystyle { \frac {\frac{b^2}{a^2} + \frac{a^2}{b^2}}{2}\ge \sqrt {\frac{b^2}{a^2} \times \frac{a^2}{b^2}} = 1 } $
$ \displaystyle { \frac {\frac{b}{a} + \frac{a}{b}}{2}\ge \sqrt {\frac{b}{a} \times \frac{a}{b}} = 1 } $

Also we apply the square mean inequality to have

$ \displaystyle {\frac{a^2 + b^2}{2} \ge \left(\frac{a+b}{2} \right)^2} $ Since a+b = 1, we have $ a^2 + b^2 \ge \frac{1}{2} $

Combining all of them we have $ \displaystyle {\left( a + \frac{1}{a} \right)^2 + \left(b + \frac{1}{b} \right)^2 \ge 12 \frac{1}{2} = \frac{25}{2}} $


Discussion

Some theoretical background.

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