$\frac{a}{b}+\frac{b}{a}>2$, same answer but different solutions?

Question

I can find the answer (correct?) to the following $$ \frac{a}{b}+\frac{b}{a}>2 \tag 1 $$ However according to the book my solution is "wrong". Why is that? Which lines are wrong? \begin{align} a^2+b^2&>2ab\tag 2\\ a^2-ab&>ab-b^2\tag 3\\ a(a-b)&>b(a-b)\tag 4\\ a&>b \tag 5 \end{align}

I also tried the following and found the same answer: \begin{align} a^2+b^2-2ab&>0\tag 6\\ (a-b)^2&>0\tag 7\\ a>b \tag 8 \end{align}

What is the difference between the solutions?

Answer 1

For ab<0 the given inequality is not true then assume $a,b>0$.

Note that from line $(4)$

$$a(a-b)>b(a-b)$$

dividing both sides by $a-b\neq 0$ we obtain

• $a>b$ when $a-b>0$

• $a<b$ when $a-b<0$

that is $a\neq b$ which is the same result we obtain from line $(6)$, indeed

$$a^2+b^2>2ab \iff a^2+b^2-2ab>0\iff(a-b)^2>0$$

wich is true for $a-b\neq 0 \iff a\neq b$,

Therefore your derivations are wrong in line $(5)$ and line $(8)$ which should be $a\neq b$.

As an alternative by Rearrangement inequality for $(a,b)$ and $(1/a,1/b)$ we have that

$$\frac{a}{b}+\frac{b}{a}\ge \frac{a}{a}+\frac{b}{b}=2$$

and equality holds if and only if $a=b$ therefore the given inequality holds for $a\neq b$.

Answer 2

First solution is wrong here

\begin{align} a(a-b)&>b(a-b)\tag 4\\ a&>b \tag 5 \end{align} You can divide with $a-b$ if you know it is $>0$. If you want to divide with $a-b$ you should assume that $a>b$ (which you may because of simmetry).

Answer 3

Im step 5 you use that $a-b$ is positive. So, also consider the negative variant here. I don't see why your Equation 8 is a logical consequence. Why don't you also include $a<b$ here?