A sum of squares of a + 1/a, via Jensen

Convexity. $f(x)=x^2+2+\frac1{x^2}$, so
$$f''(x)=2+\frac6{x^4}>0\quad(x>0).$$

Jensen. For a convex $f$: $\frac{f(a)+f(b)+f(c)}3\ge f\left(\frac{a+b+c}3\right)=f\left(\frac13\right)$. Here $f\left(\frac13\right)=\left(\frac13+3\right)^2=\frac{100}9$. Hence
$$f(a)+f(b)+f(c)\ge3\cdot\frac{100}9=\frac{100}3.$$

Equality iff $a=b=c=\frac13$.

Remark. An alternative is the tangent line trick at $\frac13$.

Lesson. If $\sum f(a_i)$ has a *fixed sum* $\sum a_i$ and $f''\ge0$, Jensen gives the minimum at the equal point; $f''\le0$ gives the maximum.