Answer
$\int_0^1 \sec x dx \geq \dfrac{7}{6}$
Work Step by Step
As we are given that $f(x)=\sec x \geq 1+\dfrac{x^2}{2}$ for $x \geq 0$
When the function $f(x)$ has minimum and maximum values defined on $[p,q]$ then the maximum-minimum inequality tells that
$min f(q-p) \leq \int_p^q f(x) dx \leq max f(q-p)$
Here, we have $ \int_0^1 \sec x dx \geq \int_0^1 (1+\dfrac{x^2}{2}) dx$
$\implies \int_0^1 \sec x dx \geq \int_0^1 dx +(\dfrac{1}{2}) \int_0^1 x^2 dx$
This gives: $\int_0^1 \sec x dx \geq 1+\dfrac{1}{6}$
Hence, $\int_0^1 \sec x dx \geq \dfrac{7}{6}$
