Let $a>b$ and $ab=1$ show that $\frac{a^2+b^2}{a-b} \geq 2\sqrt{2}$

Question

Let $a>b$ and $ab=1$. Show that $$\frac{a^2+b^2}{a-b} \geq 2\sqrt{2}$$

My attempt: $a-b>0$
$$a^2+b^2\geq 2\sqrt{2}(a-b)$$ $$\frac{a^2+b^2}{2}\geq \sqrt{2}(a-b)$$ By AG inequality we know that $$\frac{a^2+b^2}{2}\geq \sqrt{a^2b^2}=ab=1$$ But this isn't very helpful.

Answer 1

You may proceed as follows using AM-GM:

$$\frac{a^2+b^2}{a-b} = \frac{a^2+b^2 - 2ab + 2}{a-b} = \frac{(a-b)^2 + 2}{a-b}= (a-b)+ \frac{2}{a-b} \stackrel{AM-GM}{\geq}2\sqrt{2}$$

Answer 2

Consider the map$$\begin{array}{rccc}f\colon&(1,\infty)&\longrightarrow&\mathbb R\\&a&\mapsto&\dfrac{a^2+\frac1{a^2}}{a-\frac1a}.\end{array}$$Then$$f'(a)=\frac{\left(a^2+1\right) \left(a^4-4 a^2+1\right)}{a^2\left(a^2-1\right)^2}.$$You can easily deduce from this that the minimum of $f$ is attained when $a=\sqrt{2+\sqrt3}$. But $f\left(\sqrt{2+\sqrt3}\right)=2\sqrt2$.

Answer 3

Let $x=a-b>0$, then we have to prove $$(b+x)^2+b^2 \geq 2x\sqrt{2}$$ or $$2b^2+2bx+x^2\geq 2x\sqrt{2}$$ Since $1=ab = b^2+bx$ we have to prove$$2+x^2\geq 2x\sqrt{2}$$

which is the same as $$(x-\sqrt{2})^2\geq 0$$ and we are done.

Answer 4

Without applying any of the inequality identities, you can do it the tedious way using calculus from $b=1/a$ to give $$\frac{a^2+b^2}{a-b}=\frac{a^2+1/a^2}{a-1/a}=\frac{a^4+1}{a(a^2-1)}$$ The stationary point of this occurs when the derivative is zero; that is, $$\frac{4a^4(a^2-1)-(a^4+1)(3a^2-1)}{a^2(a^2-1)^2}=0$$ so $$4a^6-4a^4-3a^6+a^4-3a^2+1=a^6-3a^4-3a^2+1=0.$$ Note that $a\ne0,\pm1$ as $a>b$. The roots of the above are equivalent to the square roots of the roots of the corresponding cubic; that is, $$a^2=-1,2-\sqrt3,2+\sqrt3$$ so $a=\sqrt{2+\sqrt3}$ and $b=\sqrt{2-\sqrt3}$ as $a>b$. It is possible to find the second derivative to show that it is a minimum. Hence $$\frac{a^2+b^2}{a-b}\ge\frac{2+\sqrt3+2-\sqrt3}{\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}}=2\sqrt2$$ as required.