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I was thinking of using induction, but i am not sure how to use induction when I have two variables $x$ and $y$. Can I just prove it for any $x$ and make $y$ a certain real number like $2$?

For real numbers $x, y, z$, and $x, y \geq 0$, prove that $$\frac{x+y}{2} \geq\sqrt{xy}.$$

A. Goodier
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tskgreen
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2 Answers2

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What about $z$? It is not present in your inequality, which seems to be the classical AM-GM inequality. We have $(\sqrt{x}-\sqrt{y})^2\ge 0$, so $$x+y-2\sqrt{xy}\ge 0$$ which, after small rearrangment gives us $$\frac{x+y}{2}\ge\sqrt{xy}.$$

szw1710
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  • z is part of a different part of the problem sorry. There is no z here. But where are you getting this (sqrtx-sqrty)^2 is greater than or equal to 0?? – tskgreen Dec 17 '17 at 21:03
  • Oh, because any square power is non-negative. This is trivial. – szw1710 Dec 17 '17 at 21:48
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It's the classical AM-GM inequality which can be generalized for n variables:

$$\frac{\sum x_i}{n}\ge \prod{x_i}$$

and also it can be generalized for any exponent defining the p-Mean:

$$M(p)=\sqrt[p]\frac{\sum x_i^p}{n}$$

it can be shown that:

$$p_1\geq p_2 \iff M(p_1) \geq M(p_2)$$

Note that: $AM = M(1)$ and $GM=M(p) \quad p\to 0$

user
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  • What is AM-GM inequality – tskgreen Dec 17 '17 at 21:06
  • @tskgreen take a look here https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means and here https://artofproblemsolving.com/wiki/index.php?title=Root-Mean_Square-Arithmetic_Mean-Geometric_Mean-Harmonic_mean_Inequality – user Dec 17 '17 at 21:11