### Warning: This is an "anti-demo" of dangerous practices that lead to bad proofs. This is an example of what **_not_** to do.
Let $$a,b$$ be numbers such that $$ab \ge 0$$. "Prove" that $$a+b \ge 2\sqrt{ab}$$.
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### "Proof"
$$a + b \ge 2\sqrt{ab}$$
$$a^2 + b^2 + 2ab \ge 4ab$$
$$a^2 + b^2 - 2ab \ge 0$$
$$(a - b)^2 \ge 0$$
which is true because squares of numbers are always non-negative.
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### Discussion of what went wrong
1. The statement isn't even true.\
Consider $$a = -2$$ and $$b = -8$$.\
Then $$ab = 16 > 0$$ but $$a + b = -10$$ and $$2\sqrt{ab} = 8$$ since $$\sqrt{16} = 4$$.
2. Common fallacy: trying to prove $$p$$ by showing that $$p \Rightarrow T$$. This is "backwards" from what we actually want in a proof. In a proof, we start with a statement that is (assumed to be) true, and then prove $$p$$ ($$q \Rightarrow p$$ where $$q$$ is (assumed to be) true).\
Valid: $$1 = 1 \Rightarrow 1 + 1 = 1 + 1 \Rightarrow 2 = 2$$. This proves that $$2 = 2$$.\
Invalid: $$-1 = 1 \Rightarrow (-1)^2 = 1^2 \Rightarrow 1 = 1$$. This does **not** imply that $$-1 = 1$$.
3. Manipulating inequalities into other inequalities can be very dangerous as it allows mistakes to go unnoticed.\
$$a + b \ge 2\sqrt{ab}$$
and
$$a^2 + b^2 + 2ab \ge 4ab$$
are **not** equivalent when $$a + b < 0$$.\
Prefer "chaining" (in)equalities, so we know that every (in)equality in the chain is a true statement.\
Example of a correct statement: Let $$a,b \ge 0$$. Then $$a + b \ge 2\sqrt{ab}$$.\
**Proof**:\
$$\begin{align*}
a + b &= \sqrt{(a + b)^2} & a,b \ge 0, \text{ so } a+b \ge 0\\
&= \sqrt{a^2 + b^2 + 2ab} \\
&= \sqrt{a^2 + b^2 - 2ab + 4ab} \\
&= \sqrt{(a - b)^2 + 4ab} \\
&\ge \sqrt{4ab} & (a - b)^2 \ge 0 \\
&= 2\sqrt{ab}.
\end{align*}$$
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