The geometric mean and the arithmetic mean of two positive numbers

Question

I don't have really any idea about it.

Use the inequality: $$ \alpha \beta \;\leq\; \int\limits_{0}\limits^{\alpha}{{t}^{p-1}}\,dt \,+ \int\limits_{0}\limits^{\beta}{{u}^{q-1}}\,{du} \;=\; \frac{\alpha^p}{p} + \frac{\beta^q}{q}, $$ where $\frac{1}{q}+\frac{1}{p}=1$, to show that the geometric mean of two positive numbers does not exceed the arithmetic mean.

I just know that the geometric mean of two numbers is a special type of average where we multiply the numbers together and take a square root.

On the other hand, the arithmetic mean of two numbers is the sum of the numbers and then dividing the sum by two.

Even that, I couldn't use the inequality above to show what is wanted.

Any ideas?

.... with my great appreciation.

Answer 1

Hint: $\alpha\beta$ (the leftmost term) is the geometric mean of which two quantities? What value of $p$ and $q$ do you need to make those same two quantities appear in $\frac{{\mathit{\alpha}}^{p}}{p}\mathrm{{+}}\frac{{\mathit{\beta}}^{q}}{q}$ (the rightmost term)?

Answer 2

There must be some missing assumptions here (beyond assuming the variables are all positive), because the inequality

$$\alpha\beta\le{\alpha^p\over p}+{\beta^q\over q}$$

is most decidedly not true in general. Take $p=q=1$, for example. No one would claim that $\alpha\beta\le\alpha+\beta$ for all (positive) $\alpha$ and $\beta$.

The inequality is, of course, true, when you set $p=q=2$, because it essentially is the AGM inequality in that case. But it's kind of a weird assignment to prove a true statement by starting with something that's false in general.

Answer 3

It is a problem form keryszig functional analysis. Here, $$\alpha\beta\le{\alpha^p\over p}+{\beta^q\over q} $$ $$\frac{1}{q}+\frac{1}{p}=1$$ where $ p>1$ , From the definiton of geometric mean

$$\Biggl({\prod_{i=0}^n x_i}\Biggr)^{1/n} = \sqrt[n]{x_1.x_2....x_n}$$ and from the definition of arithmetic mean $$ {1 \over n} \sum_{i=0}^n x_i = {{x_i + x_2+...+x_n}\over n} $$ Now, $\alpha > 0, \beta>0 $ . And taking $ p =2$ , We get $ q= 2$

$$\alpha\beta \le {\alpha^2\over 2} + {\beta^2\over 2} $$ $$2\alpha\beta \le {\alpha^2} + {\beta^2}$$ Adding $2\alpha\beta $ on both side we get, $$ 4\alpha\beta \le {\alpha^2} + {\beta^2} + 2\alpha\beta$$ $$4\alpha\beta \le {(\alpha + \beta)}^2$$ Now taking square root on both sides $$2\sqrt{\alpha\beta} \le \alpha + \beta$$ $$\sqrt{\alpha\beta} \le {(\alpha + \beta)\over 2}$$ Where $\sqrt{\alpha\beta}$ is geometric mean for $\alpha $ and $\beta$ and ${(\alpha + \beta)\over 2}$ is airthmetic mean. Hence the geometric mean of two positive numbers does not exceed the arithmetic mean .