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