direct proofs of inequalities

Question

I completed an entire chapter set on direct proofs only to find my teacher said reject my answers due to a false method. what i did was assume the problem was true and then solved it as such and then wrote it backwards. Her exact words were you should only use this method to search for hints to achieve a direct proof, as it will help you gain the skills needed for these types of questions. Further, when attempting to solve these only work on one side - I did not know what this meant and still don't really: Also, how are you meant to prove a question if you are only working with one side?

Once taking this advice i found myself stuck on how to presume on the questions i had previously solved, as i would be trying to find out how to reduce the LHS to an acceptable answer only to which i found myself even more confused about how to pursue these questions. Not to my aid the tutorials on YouTube are not good, Hence, my asking for your help for essentially what are the rules when proving inequalities - and to clarify it is only the inequalities that i find difficult - when can you add, divide...etc, when can subtract from both side - LHS and RHS - etc...

if you could please answer me the basics that would be helpful!

thanks to anyone who actually took the time to read that lol !

for example:

Show that for all a, b∈R, $$(\frac{a^2 + b^2}{2})⩾(\frac{a+b}{2})^2$$

which i answered as:

$$\frac {a^2 + b^2}{ab}-2 ⩾0$$

$$\frac {a^2 + b^2}{ab} ⩾2$$

$$ a^2 + b^2 ⩾2ab $$

$$ 2a^2 + 2b^2 ⩾ a^2 + 2ab + b^2 $$

$$\frac {a^2+b^2}{2}⩾ \frac{a^2 + 2ab + b^2}{4}$$

$$\frac {a^2+b^2}{2}⩾ \frac{(a+b)^2}{4}$$

Answer 1

Your first step is wrong for $ab\leq0.$

The right proof is the following.

We need to prove that $$2(a^2+b^2)\geq(a+b)^2$$ or $$a^2-2ab+b^2\geq0$$ or $$(a-b)^2\geq0,$$ which is true.

Id est, $$\frac{a^2+b^2}{2}\geq\left(\frac{a+b}{2}\right)^2$$ is true and we are done because in all lines we have equivalent statements.

Answer 2

You have proved the inequality, provided that $$ \frac{a^2+b^2}{ab}-2\ge0 $$ which is generally false and even meaningless when $a=0$ or $b=0$.

Note that, even for $ab\ne0$, you can't go from $$ a^2+b^2\ge2ab $$ to $$ \frac{a^2+b^2}{ab}\ge2 $$ Try $a=1$ and $b=-1$, for instance.

A correct derivation would be \begin{gather} (a-b)^2\ge0 \\ a^2+b^2-2ab\ge0 \\ a^2+b^2\ge2ab \\ 2a^2+2b^2\ge a^2+2ab+b^2 \\ \frac{2(a^2+b^2)}{4}\ge\frac{(a+b)^2}{4} \\ \frac{a^2+b^2}{2}\ge\frac{(a+b)^2}{2^2} \\ \frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2 \end{gather}