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}}.