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| − | AoPS Wiki users, ignore this page. I'm using my User Talk to explain proof writing to friends.
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| − | == Proof 1: Inequalities ==
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| − | 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.)
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| − | === Explanation ===
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| − | 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):
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| − | <cmath>\begin{align*}
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| − | \frac{x + y}{2} \geq \sqrt{xy} \\
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| − | x + y \geq 2 \sqrt{xy} \\
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| − | (x + y)^2 \geq 4xy \\
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| − | x^2 + 2xy + y^2 \geq 4xy \\
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| − | x^2 - 2xy + y^2 \geq 0 \\
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| − | (x - y)^2 \geq 0.
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| − | \end{align*}</cmath>
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| − | 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:
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| − | === Bad Proof ===
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| − | 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>
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| − | '''Why is this proof bad?'''
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| − | * '''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.
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| − | * '''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''.
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| − | * '''Not Enough Space''': A little more space would make this proof easier to read. Important equations should be given their own line.
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| − | === Good Proof ===
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| − | 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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