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Inequality Problem (Tomato subjective 83)

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

Solution: \({\displaystyle{\left(a + {\frac{1}{a}}\right)^2 + \left(b + {\frac{1}{b}}\right)^2 {\ge} {\frac{25}{2}}}}\)
\({\displaystyle{\Leftrightarrow}}\) \({\displaystyle{a^2 + b^2 + {\frac{1}{a^2}} + {\frac{1}{b^2}} {\ge} {\frac{17}{2}}}}\) … (i)
Now \({\displaystyle{a + b = 1}}\) \({\displaystyle{\Rightarrow}}\) \({\displaystyle{a^2 + b^2 + 2ab = 1}}\)
\({\displaystyle{\Rightarrow}}\) \({\displaystyle{a^2 + b^2 {\ge} {\frac{1}{2}}}}\) … (ii)
From (i) & (ii) we get to prove \({\displaystyle{{\frac{1}{a^2}} + {\frac{1}{b^2}} {\ge} {8}}}\)
\({\displaystyle{\Leftrightarrow}}\) \({\displaystyle{{\frac{a^2 + b^2}{a^2 b^2}} {\ge} {8}}}\)
\({\displaystyle{\Leftrightarrow}}\) \({\displaystyle{{\frac{1}{a^2 b^2}} {\ge} {4}}}\) [ as \({a^2 + b^2 {\ge} {\frac{1}{2}}}\) ]
\({\displaystyle{\Leftrightarrow}}\) \({\displaystyle{1 {\ge} {4 a^2 b^2}}}\)
\({\displaystyle{\Leftrightarrow}}\) \({\displaystyle{1 {\ge} (a + b)^2}}\)
Now this follows directly from the given condition \({\displaystyle{a + b = 1}}\).

July 28, 2015
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