Difference between revisions of "User talk:Etmetalakret"

Revision as of 19:47, 1 April 2023

AoPS Wiki users, ignore this page. I'm using my User Talk to explain proof writing to friends.

Proof 1: Inequalities

The well-known Trivial Inequality states that if $x$ is a real number, then $x^2 \geq 0$ . Prove that if $x$ and $y$ are nonnegative real numbers, then $\frac{x + y}{2} \geq \sqrt{xy}.$ (Sidenote: this is a very different kind of inequality problem than you're used to. In school, we find when inequalities are true; here, we're showing it's always true.)

Explanation

I found the proof by working backwards; I started with the desired result, and connected it to something true. Here is the wall of equations on my page (sadly I can't get them aligned): $\begin{align*} \frac{x + y}{2} \geq \sqrt{xy} \\ x + y \geq 2 \sqrt{xy} \\ (x + y)^2 \geq 4xy \\ x^2 + 2xy + y^2 \geq 4xy \\ x^2 - 2xy + y^2 \geq 0 \\ (x - y)^2 \geq 0. \end{align*}$ Because the left-hand side of this equation is a perfect square, this is actually the Trivial Inequality in disguise. The desired inequality is therefore implied by a true result. Really understand and grasp how I derived this before you read the following proofs:

Bad Proof

I start out with $\frac{x + y}{2} \geq \sqrt{xy}.$ Multiply the inequality by $2$ and square it, $(x + y)^2 \geq 2 \sqrt{xy}$ . Letting our algebra go on autopilot, $x^2 + 2xy + y^2 \geq 4xy$ and $x^2 - 2xy + y^2 \geq 0$ , so $(x - y)^2 \geq 0$ . This is true by Trivial Inequality, which completes the proof. $\square$

Why is this proof bad?

Written Backwards: We must always write proofs like: true result $\implies$ desired result. However, the proof is written backwards so that the desired result $\implies$ true result. The Trivial Inequality should be at the start, not the end.
Informal Word Choice: Please don't use the phrase "algebra autopilot" in a proof, and don't write sentences with no verbs (see the "Multiply the inequality by $2$ and square it"). Also, don't use "I," although "we" is totally acceptable.
Not Enough Space: A little more space would make this proof easier to read. Important equations should be given their own line.

Good Proof

By the Trivial Inequality, we have that $(x - y)^2 \geq 0.$ Factoring this inequality returns $x^2 - 2xy + y^2 \geq 0$ . We add $4xy$ to both sides and factor to get $(x + y)^2 \geq 4xy$ . Note that because $x$ and $y$ are nonnegative, both sides are nonnegative; we may therefore take the square root of the inequality, which yields $x + y \geq 2 \sqrt{xy}.$ Finally, dividing both sides by $2$ gives the desired inequality of $(x + y) / 2 \geq \sqrt{xy}$ . $\square$

@@ Line 1: / Line 1: @@
-🦍Hello🦍
+AoPS Wiki users, ignore this page. I'm using my User Talk to explain proof writing to friends.
-This is my user talk page, where you can leave me a message.
+== Proof 1: Inequalities ==
+The well-known '''Trivial Inequality''' states that if <math>x</math> is a real number, then <math>x^2 \geq 0</math>. Prove that if <math>x</math> and <math>y</math> are nonnegative real numbers, then <cmath>\frac{x + y}{2} \geq \sqrt{xy}.</cmath> (Sidenote: this is a very different kind of inequality problem than you're used to. In school, we find ''when'' inequalities are true; here, we're showing it's ''always'' true.)
-Maybe you wanna offer feedback on something I've done, tell me that a page I rewrote is garbage, complain that I edit my user page too much (I do), or even just say hello — whatever you leave here, I'll be more than happy to read and respond to it.
+=== Explanation ===
+I found the proof by ''working backwards''; I started with the desired result, and connected it to something true. Here is the wall of equations on my page (sadly I can't get them aligned):
+<cmath>\begin{align*}
+\frac{x + y}{2} \geq \sqrt{xy} \\
+x + y \geq 2 \sqrt{xy} \\
+(x + y)^2 \geq 4xy \\
+x^2 + 2xy + y^2 \geq 4xy \\
+x^2 - 2xy + y^2 \geq 0 \\
+(x - y)^2 \geq 0.
+\end{align*}</cmath>
+Because the left-hand side of this equation is a perfect square, this is actually the Trivial Inequality in disguise. The desired inequality is therefore implied by a true result. Really understand and grasp how I derived this before you read the following proofs:
-You can also just call me James if etmetalakret (pronounced et-met-ala-kret) is too much of a hassle to type
+=== Bad Proof ===
+I start out with <math>\frac{x + y}{2} \geq \sqrt{xy}.</math> Multiply the inequality by <math>2</math> and square it, <math>(x + y)^2 \geq 2 \sqrt{xy}</math>. Letting our algebra go on autopilot, <math>x^2 + 2xy + y^2 \geq 4xy</math> and <math>x^2 - 2xy + y^2 \geq 0</math>, so <math>(x - y)^2 \geq 0</math>. This is true by Trivial Inequality, which completes the proof. <math>\square</math>
+'''Why is this proof bad?'''
+* '''Written Backwards''': We must always write proofs like: true result <math>\implies</math> desired result. However, the proof is written backwards so that the desired result <math>\implies</math> true result. The Trivial Inequality should be at the ''start'', not the end.
+* '''Informal Word Choice''': Please don't use the phrase "algebra autopilot" in a proof, and don't write sentences with no verbs (see the "Multiply the inequality by <math>2</math> and square it"). Also, don't use "I," although ''"we" is totally acceptable''.
+* '''Not Enough Space''': A little more space would make this proof easier to read. Important equations should be given their own line.
+=== Good Proof ===
+By the Trivial Inequality, we have that <cmath>(x - y)^2 \geq 0.</cmath> Factoring this inequality returns <math>x^2 - 2xy + y^2 \geq 0</math>. We add <math>4xy</math> to both sides and factor to get <math>(x + y)^2 \geq 4xy</math>. Note that because <math>x</math> and <math>y</math> are nonnegative, both sides are nonnegative; we may therefore take the square root of the inequality, which yields <cmath>x + y \geq 2 \sqrt{xy}.</cmath> Finally, dividing both sides by <math>2</math> gives the desired inequality of <math>(x + y) / 2 \geq \sqrt{xy}</math>. <math>\square</math>

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