<s>[INST] Minimizing $\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}$ over positive reals with $a+b=1$. Why is the minimum not $18$? 
If $a,b \in R^+$ such that $a+b=1$, then find the minimum value of $$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}$$

We can write $$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}=\left(a+\frac{a+b}{a}\right)^{2}+\left(b+\frac{a+b}{b}\right)^{2}=\left(2+\frac{a}{b}\right)^{2}+\left(2+\frac{b}{a}\right)^{2}$$
Using $Q.M\geq A.M\geq G.M$ we have
$$\sqrt{\frac{\left(2+\frac{a}{b}\right)^{2}+\left(2+\frac{b}{a}\right)^{2}}{2}} \geq \frac{2+\frac{a}{b}+2+\frac{b}{a}}{2}=2+\frac{\frac{a}{b}+\frac{b}{a}}{2} \geqslant 2+1=3$$
So
$$\begin{array}{l}
\left(2+\frac{a}{b}\right)^{2}+\left(2+\frac{b}{a}\right)^{2} \geqslant 18 \\
\Rightarrow \quad\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2} \geqslant 18
\end{array}$$

But in an alternate approach i got the correct minimum as $12.5$

 [/INST] Another variation on the root-mean-square inequality (or Cauchy-Schwarz with $c = d = 1$) is:
$$\left(x + \frac{1}{x} \right)^2 + \left(1-x+\frac{1}{1-x} \right)^2 \ge \frac{\left(x + \frac{1}{x} +1-x+\frac{1}{1-x} \right)^2}{2} \tag{$0 \le x \le 1$}$$
$$= \frac{1}{2} \left(1 + \frac{1}{x} + \frac{1}{1-x} \right)^2 = \frac{1}{2} \left(1 + \frac{1}{x(1-x)} \right)^2.$$
Now this expression is minimised when $x(1-x)$ is maximised, which occurs halfway between the roots $x = 0,\ 1$. Thus the minimum value is $\frac{1}{2} \left(1 + \frac{1}{\frac{1}{2}(1-\frac{1}{2})} \right)^2 = \boxed{\frac{25}{2}}.$
</s>