# 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 $$12.5$$

• why is $a+\frac{a+b}{a} = 2+\frac{a}{b}$ or $2+\frac{b}{a}$? Nov 28, 2020 at 7:32

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