Cfrgtkky

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

My attempt:
$a-b>0$
$$a^2+b^2geq 2sqrt{2}(a-b)$$
$$frac{a^2+b^2}{2}geq sqrt{2}(a-b)$$
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.

You may proceed as follows using AM-GM:

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

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

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 $ane0,pm1$ 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 $$frac{a^2+b^2}{a-b}gefrac{2+sqrt3+2-sqrt3}{sqrt{2+sqrt3}-sqrt{2-sqrt3}}=2sqrt2$$ as required.

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

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