# Proof that arithmetic mean is greater than geometric mean? [duplicate]

I have to prove that $\frac{x + y}{2}> \sqrt{xy}$ algebraically for any $x,y \in \mathbb{R}$ such that $x,y \ge 0$ and $x\ne y.$

I'm fairly confused as to how to solve this problem algebraically, where would be a good place to start?

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## marked as duplicate by David H, Fundamental, Hakim, Thomas Klimpel, Thomas AndrewsJul 10 '14 at 21:59

Dit you write it correctly? For if we have $x=8$ and $y=2$ we obtain $(x+y)/x < \sqrt{xy}$... –  johannesvalks Jul 10 '14 at 20:15
@johannesvalks sorry about that! Don't know how the x got in there! I meant (x + y) / 2 –  Dunka Jul 10 '14 at 20:18
Try to square both terms of the inequality. –  mfl Jul 10 '14 at 20:20
The simplest derivation is to notice that $(\sqrt{x} - \sqrt{y})^2 = x + y - 2\sqrt{xy}$ so $\frac{x+y}{2} > \sqrt{xy} \iff (\sqrt{x} - \sqrt{y})^2 > 0$ –  Winther Jul 10 '14 at 21:11
If interested, there is a nice proof of the fact that $\textbf{harmonic mean}$ $\leq$ $\textbf{geometric mean}$ $\leq$ $\textbf{arithmetic mean}$ in Proofs from the book by Aigner & Ziegler –  user128779 Jul 10 '14 at 21:58

I think you want to show

$$\frac{x+y}{2} > \sqrt{xy}.$$

Write

$$\begin{eqnarray} \frac{x+y}{2} =\\ \sqrt{ \left( \frac{x + y}{2} \right)^2 } &>& \sqrt{ \left( \frac{x + y}{2} \right)^2 - \left( \frac{x - y}{2} \right)^2 }\\ && = \sqrt{ \left( \frac{x + y}{2} + \frac{x - y}{2}\right) \left( \frac{x + y}{2} - \frac{x - y}{2} \right) }\\ && = \sqrt{xy}, \end{eqnarray}$$ whence

$$\frac{x+y}{2} > \sqrt{xy}.$$

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There's also a nice geometric proof.

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Woolfit: Wow a nice proof - I always like visual proof!!!! Voteup (this should be the best answer selected imo...) –  johannesvalks Jul 10 '14 at 20:41
+1. There is an old saying: "An image is better than 1000 words". –  Felix Marin Jul 10 '14 at 20:58

Multiply by two to get $$x+y\geq2\sqrt{xy}$$ then square to get $$(x+y)^2=x^2+y^2+2xy\geq4xy$$ then substract $4xy$ from both sides to get $$x^2-2xy+y^2=(x-y)^2\geq0$$ this is obvious since the square of a real number is never negative.

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But then you should also explain why you can make "if...then..." go in the opposite direction here. I.e. in effect you're saying "If $(x+y)/2\ge\sqrt{xy}$ then $(x+y)^2\ge4xy$", but what you need is "If $(x+y)^2\ge4xy$ then $(x+y)/2\ge\sqrt{xy}$". And so on. ${}\qquad{}$ –  Michael Hardy Jul 10 '14 at 21:02

More two solutions:

i) \begin{align} (\sqrt{x} - \sqrt{y})^2 &\geq 0 \\ x - 2\sqrt{xy} + y &\geq 0 \\ x + y &\geq 2\sqrt{xy} \\ \frac{x + y}{2} &\geq \sqrt{xy} \end{align}

ii) If $x = y$, it's trivial. Suppose $0 < y < x$ and fix $y$. Define $$f(x) = \frac{x + y}{2} - \sqrt{xy}$$ Take the derivative: $$f'(x) = \frac{1}{2} - \frac{\sqrt{y}}{2\sqrt{x}} > 0$$ Then, since $f(0) = y/2 > 0$ and $f$ is increasing, we get the desired inequality.

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solution i is nice. How did you know to do that though? –  Dunka Jul 10 '14 at 21:25
I sort of "reverse engineered" it. I considered that because I knew it would show up $x, y$ and $\sqrt{xy}$. –  Ivo Terek Jul 10 '14 at 21:35