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Answer: x+y ≥ 2√(xy) ⇔ (x+y)^2 ≥ 4xy ⇔ x^2+y^2+2xy ≥ 4xy ⇔ (x−y)^2 ≥ 0, which is true. Equality holds when x=y.

I've solved it until (x−y)^2 ≥ 0, but I don't understand "which is true". If y is larger than x won't it not be true? e.g. 14 - 16 = -2. (I get this now!)

Also, it means x is larger than or equal to y. How do we know equality holds when x=y and not when x is larger than y? (Still don't get this)

Thank you.

user30200
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1 Answers1

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You are missing the forest for the trees.

It doesn't matter if $x \ge y$ or $y\ge x$ because $(x-y)^2$ is a square. And if $x \ne y$ then $(x-y)^2 = (y-x)^2 > 0$. And $(x-y)^2 = 0 \iff x-y = 0 \iff x=y$.

There are only three options.

  1. $x > y$. Then $x-y > 0$. And $(x-y)^2 > 0$. For example. If $x = 16$ and $y=14$ then $x+y= 30$ and $2\sqrt{xy} = 2\sqrt{14*16} = 8\sqrt {14} < 30$. And $(x-y)^2 = (16-14)^2 = 2^2 = 4 > 0$.

  2. $x = y$. Then $x-y = 0$. And $(x-y)^2 = 0$. For example. If $x = 16$ and $y=16$ then $x+y = 32$ and $2\sqrt{16*16}=2\sqrt{256}=2*16 = 32$. And $(x-y)^2 = (16-16)^2 = 0^2 = 0$.

and

  1. $x < y$ then $x-y < 0$ and $(x-y)^2 > 0$. For example. If $x = 14$ and $y-16$ then $x+y = 30$ and $2\sqrt{xy} = 2\sqrt{16*14} = 8\sqrt{14} < 30$. And $(x-y) = (14-16)^2 = (-2)^2 = 4 > 0$.
fleablood
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