For two positive real numbers $x$ and $y$, we can prove the AM-GM inequality between them.

$$\left(\sqrt{x}-\sqrt{y}\right)^2=x+y-2\sqrt{xy}\geq 0$$

$$\implies \frac{x+y}{2}\geq\sqrt{xy}$$

Can this be generalized easily to the case of more numbers?

  • $\begingroup$ This can be done, using some kind of a "non-ascending" induction. You induct on the natural numbers, but not in a conventional manner. You kind of go up in powers of $2$, and descend for non-powers of $2$ from the nearest larger power of $2$. Otherwise, you can cheat and use Lagrange multipliers. $\endgroup$ – астон вілла олоф мэллбэрг Sep 24 '16 at 3:35
  • $\begingroup$ @Arturo don Juan Read about the Jensen's inequality. $\endgroup$ – Michael Rozenberg Sep 24 '16 at 4:13

Yes and it can be proved in a simple way using Langrange multipliers.

Let $x_1, ... x_n$ be real numbers such that $x_1 + ... + x_n = S$, where $S$ is some fixed number. Show that the product $x_1...x_n$ is minimized when $x_1 = x_2 =...=x_n$ and conclude.


If $n=2^k$, where $k\in\mathbb N$, so we can use $\frac{x+y}{2}\geq\sqrt{xy}$ for non-negatives $x$ and $y$.

If $n>2^k$ we can use a similar to the following reasoning.

We'll prove that $\frac{x+y+z}{3}\geq\sqrt[3]{xyz}$ for non-negatives $x$, $y$ and $z$.

Indeed, by AM-GM for four variables we obtain: $$\frac{x+y+z+\sqrt[3]{xyz}}{4}\geq\sqrt[4]{xyz\sqrt[3]{xyz}}=\sqrt[3]{xyz}$$ and we are done!

For five non-negative variables we have $$\frac{a+b+c+d+e+3\sqrt[5]{abcde}}{8}\geq\sqrt[8]{abcde\sqrt[5]{a^3b^3c^3d^3e^3}}=\sqrt[5]{abcde}$$ which gives $$\frac{a+b+c+d+e}{5}\geq\sqrt[5]{abcde}$$


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