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 ${\displaystyle{a + b = 1}}$.