Proving inequality $(a+\frac{1}{a})^2 + (b+\frac{1}{b})^2 \geq \frac{25}{2}$ for $a+b=1$ If $a, b$ are positive real numbers and $a+b = 1$, prove that :
$$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}$$
I can see that the value $\frac{25}2$ is attained for $a=b=\frac12$. But I do not know how to show that this is the minimal possible value.
Thank you.

Let $f(p,q)=\left(p+\frac{1}{p}\right)^2+\left(q+\frac{1}{q}\right)^2$.
Using Lagrange multiplier's method:
$$L=\left(p+\frac{1}{p}\right)^2+\left(q+\frac{1}{q}\right)^2+\lambda (1-p-q)\\
\begin{cases}L_p=2\left(p+\frac1p\right)(1-\frac1{p^2})-\lambda=0\\
L_q=2\left(q+\frac1q\right)(1-\frac1{q^2})-\lambda=0\\
L_{\lambda}=1-p-q=0\end{cases} \Rightarrow p=q=\frac12$$
SOC:
$$L_{pp}=2\left(1-\frac1{p^2}\right)^2+2\left(p+\frac1p\right)\cdot \frac1{p^3}>0\\
L_{pq}=0\\
\Delta=L_{pp}L_{qq}-L_{pq}^2=L_{pp}^2>0$$
Hence: $f(\frac12,\frac12)=12.5$ is minimum.
