This is a subjective problem from TOMATO based on inequality.
Problem: Inequality Problem
If $ {\displaystyle{a}}$ and $latex {\displaystyle{b}}$ 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}}$.
