Vtyjkyuk

I can find the answer (correct?) to the following
$$
fracab+fracba>2 tag 1
$$
However according to the book my solution is "wrong". Why is that? Which lines are wrong?
beginalign
a^2+b^2&>2abtag 2\
a^2-ab&>ab-b^2tag 3\
a(a-b)&>b(a-b)tag 4\
a&>b tag 5
endalign

I also tried the following and found the same answer:
beginalign
a^2+b^2-2ab&>0tag 6\
(a-b)^2&>0tag 7\
a>b tag 8
endalign

What is the difference between the solutions?

First solution is wrong here

beginalign
a(a-b)&>b(a-b)tag 4\
a&>b tag 5
endalign
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).

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 $a-bneq 0$ we obtain

• $a>b$ when $a-b>0$




• $a<b$ when $a-b<0$

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

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

wich is true for $a-bneq 0 iff aneq b$,

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

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

$$fracab+fracbage fracaa+fracbb=2$$

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

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?