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 $18$?

Question

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 $12.5$

Answer 1

The issue is in the first line of simplification. $$a+\frac{a+b}{a}=a+1+\frac{b}{a}\neq 2+\frac{b}{a}.$$ If you make the correct simplification, you can continue $$\sqrt{\frac{\left(a+1+\frac ba\right)^2+\left(b+1+\frac ab\right)^2}{2}}\geq \frac{a+b+2+\frac ab+\frac ba}{2}=\frac{3+\frac ab+\frac ba}{2}\geq \frac 52,$$ and get the correct result.

Answer 2

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 $\frac{1}{2} \left(1 + \frac{1}{\frac{1}{2}(1-\frac{1}{2})} \right)^2 = \boxed{\frac{25}{2}}.$