Home / ab = \frac{1}{2} \quad \Rightarrow \quad \frac{1}{a^2} + \frac{1}{b^2} = \frac{(a^2 + b^2)^2}{a^2 b^2} = \frac{1}{(\frac{1}{2})^2} = 4

ab = \frac{1}{2} \quad \Rightarrow \quad \frac{1}{a^2} + \frac{1}{b^2} = \frac{(a^2 + b^2)^2}{a^2 b^2} = \frac{1}{(\frac{1}{2})^2} = 4

February 22, 2026

$ab = \frac{1}{2} \quad \Rightarrow \quad \frac{1}{a^2} + \frac{1}{b^2} = \frac{(a^2 + b^2)^2}{a^2 b^2} = \frac{1}{(\frac{1}{2})^2} = 4$

We apply the AM-GM inequality to $\frac{1}{a^2} + \frac{1}{b^2}$:
\frac{1}{a^2} + \frac{1}{b^2} \geq \frac{2}{ab}
Given $a^2 + b^2 = 1$, the maximum of $ab$ is $\frac{1}{2}$ when $a = b = \frac{\sqrt{2}}{2}$. Then:
Therefore, the expression becomes:
+ 4 = 9
Verification at $x = \frac{\pi}{4}$ shows $\cos x = \sin x = \frac{\sqrt{2}}{2}$, yielding: