Додаток: нерівність між середніми та нерівність Коші-Буняковського для двох змінних

The AM GM Inequality is that the Arithmetic Mean is always greater than or equal to the Geometric Mean.

$AM\ge GM$

AM-GM

The arithmetic mean, or average as we call it, of the numbers $a_1,a_2,\dots,a_n$ is equal to

$$AM=\frac{a_1+a_2+\dots+a_n}{n}$$

The geometric mean is defined by

$$GM=\sqrt[n]{a_1a_2\dots a_n}$$

The AM-GM inequality states that if $a_1,a_2,\dots,a_n$ are NONNEGATIVE REALS (I can't stress this more), then $AM\ge GM$, or

$$\frac{a_1+a_2+\dots+a_n}{n}\ge\sqrt[n]{a_1a_2\dots a_n}$$

Another important fact about this inequality is it attains equality if and only if $a_1=a_2=\dots=a_n$.

You may be asking me why this is important. It gives us a way of converting sums to products and products to sums! Check this example out:

What is the maximum value of the function $x(1-x)$ on the domain $[0,1]$?

Solution: Notice that $x\ge0$ and $1-x\ge0$ because of the domain, so the product is magically turned into a sum by AM-GM:

$$\sqrt{x(1-x)}\le\frac{x+(1-x)}{2}=\frac12$$

Therefore, $x(1-x)\le\left(\frac12\right)^2=\frac14$, and it attains equality where $x=1-x$, which is at $x=\frac12$, which is in the domain. The maximum is thus $\frac14$.

What is the minimum value of $x+\frac4x$ for $x>0$?

Solution: Both terms are positive. Let's turn this sum into a product!

$$\frac{x+\frac4x}2\ge\sqrt{x\left(\frac4x\right)}=\sqrt{4}=2$$

Therefore, $x+\frac4x\le2\cdot2=4$, attaining equality at $x=\frac4x$ or $x=2$, which is in the domain. Therefore, the minimum is $4$.

Sometimes in AM-GM, we don't get such cancellations so easily, so we can force them to cancel.

What is the maximum value of the function $x(1-2x)$ on the domain $[0,\frac12]$?

Solution: With AM-GM on $x,1-2x$, we get

$$\sqrt{x(1-2x)}\le\frac{x+(1-2x)}{2}=\frac{1-x}2$$

which isn't very helpful to us here. Therefore, manipulate the function as $x(1-2x)=\frac12(2x)(1-2x)$. Since the maximum of $\frac12(2x)(1-2x)$ is just $\frac12$ times the maximum of $(2x)(1-2x)$, let's find the maximum of $(2x)(1-2x)$.

$$\sqrt{(2x)(1-2x)}\le\frac{(2x)+(1-2x)}2=\frac12$$

Therefore, $(2x)(1-2x)\le\frac14$, so $x(1-2x)\le\frac12\cdot\frac14=\frac18$.

What is the minimum value of $x^2+\frac2x$ for $x>0$?

Solution: Again, no cancellation if we apply AM-GM like this. Manipulate the sum to be $x^2+\frac1x+\frac1x$. Then AM-GM gives

$$\frac{x^2+\frac1x+\frac1x}3\ge\sqrt[3]{x^2\cdot\frac1x\cdot\frac1x}=1$$

Therefore, $x^2+\frac2x\ge3$, with equality attained where $x^2=\frac1x=\frac1x$, or $x=1$ which is in the domain.

Help, I Broke AM-GM!

What is wrong with the following applications of AM-GM?

1. $$\frac{x-2-\frac1x}3\ge\sqrt[3]{x\cdot(-2)\cdot\left(-\frac1x\right)}=\sqrt[3]{2}$$

so $x-2-\frac1x\ge3\sqrt[3]2$.

1. $$x^2+\frac{16}{x}=x^2+\frac8x+\frac8x$$

and

$$\frac{x^2+\frac8x+\frac8x}{3}\ge\sqrt[3]{x^2\cdot\frac8x\cdot\frac8x}=4$$

so $x^2+\frac{16}{x}\ge12$, with equality attained at $x^2=\frac{16}{x}$ or $x=\sqrt[3]{16}$.

1. $$\frac{x^2+\frac1{x^2}}2\le\sqrt{x^2\cdot\frac1{x^2}}=1$$

so $x^2+\frac1{x^2}\le2$.

1. $$\frac{x^2+\frac8x+\frac1x}{3}\ge\sqrt[3]{x^2\cdot\frac8x\cdot\frac1x}=2$$

so $6$ is the minimum value of $x^2+\frac9x$.

Solutions

1. AM-GM is for nonnegative reals!

2. The equality case is $x^2=\frac8x=\frac8x$!

3. The inequality is going the wrong way!

4. The equality case is $x^2=\frac8x=\frac1x$, which is impossible!

The AM-GM inequality for two variables states that for two positive real numbers $a$ and $b$, $$\dfrac{a+b}{2} \geq \sqrt{ab}.$$ The left side is called the arithmetic mean of the two numbers, and the right side is called the geometric mean of the two numbers -- thus the name, AM-GM. Both means are very useful means of obtaining some sort of average of the two numbers (no pun intended), and we see here that one of them is always bigger than the other.

Now you're probably asking... why is this true? Try doing things to the inequality and see if we can turn it into something more familiar.

First we multiply by 2. Then, squaring both sides yields $$(a+b)^2 \geq 4ab.$$ If we expand, we see that the original inequality is equivalent to $a^2+2ab+b^2 \geq 4ab$, or $a^2-2ab+b^2 \geq 0$. That can be factored, so we rewrite it as $(a-b)^2 \geq 0$.

Hey, that's familiar! That's just the Trivial Inequality. Since the AM-GM inequality in two variables boils down to the Trivial Inequality, it must be true!

(You might be a little wary of the squaring step. If you are, that's great -- since squaring both sides does not always produce an equivalent inequality. [If you're not sure why, try squaring both sides of the inequality $-3 > -4$.] But here, all is well since $a$ and $b$ are positive.)

Note that equality holds when $a = b$.

Now we turn to the Cauchy-Schwarz inequality. For two variables, it states that for positive real numbers $a_1, a_2, b_1, b_2$, we have $$(a_1^2+a_2^2)(b_1^2+b_2^2) \geq (a_1b_1 + a_2b_2)^2.$$ (You can remember it easily as "the product of the sums of squares is at least the square of the sum of products.")

Let's prove it.

Let's expand -- we get: $$a_1^2b_1^2 + a_1^2b_2^2 + a_2^2b_1^2 + a_2^2b_2^2 \geq a_1^2b_1^2 + 2a_1a_2b_1b_2 + a_2^2b_2^2.$$ A whole bunch of things cancel, so the original inequality is equivalent to $$a_1^2b_2^2 + a_2^2b_1^2 \geq 2a_1a_2b_1b_2.$$

Look familiar?

That inequality we just got is equivalent to two-variable AM-GM! To see why, we rewrite it as $$\dfrac{a_1^2b_2^2 + a_2^2b_1^2}{2} \geq \sqrt{a_1^2b_2^2 \cdot a_2^2b_1^2}.$$

Thus, we're done!

Note that equality holds when $a_1b_2 = a_2b_1$, or $\frac{a_1}{b_1} = \frac{a_2}{b_2}$.

Try some of the problems below for practice. They are a mix of AM-GM and Cauchy-Schwarz problems. Don't be surprised if you use both for one problem. Good luck.

The Arithmetic Mean-Geometric Mean inequality is an inequality that states $$\dfrac{a_1+a_2+a_3\cdots a_n}{n}\geq\sqrt[n]{a_1a_2a_3\cdots a_n}$$ This inequality has many uses.

Inequalities (Part 2)

AM-GM inequality

This is a very useful inequality. Most problems requires an application of this inequality. The inequality states that :

For all positive integers $n$ and all positive real numbers $a_1, a_2, ..., a_n$, we have

$\frac{a_1 + a_2 + ... + a_n}{n} \ge \sqrt[n]{a_1a_2a_3...a_n}$.

This is the general AM-GM inequality. I will not go through the proof here but the proof $n = 4$ and $n = 3$ are worth discussing :

We first prove AM-GM for $n = 4$. Indeed,

$a + b + c + d \ge 2(\sqrt{ab} + \sqrt{cd}) \ge 4\sqrt{\sqrt{abcd}} = 4\sqrt[4]{abcd}$.

Finally, substituting $d = \frac{a + b + c}{3}$ in the AM-GM for $4$ variables yields the AM-GM for $3$ variables. Interested readers can search the proof of the general case.

Exercise. (Solutions will not be given)

1. Prove that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3$ for all positive reals $a, b, c$.

2. Prove that $x + \frac{1}{x} \ge 2$ for all positive real $x$.

3. Prove that $\frac{x}{x^4 + y^2} + \frac{y}{y^4 + x^2} \le \frac{1}{xy}$.

In the next part we will look at advanced applications of AM-GM.

Solving inequalities is a very useful skill in algebra.

Let us start with the trivial inequality. The trivial inequality states that $x^2\geq0$ for real $x$. This seemingly obvious inequality is the basis for many important inequalities. One, for example, is the arithmetic mean and geometric mean inequality, also know as AM-GM.

Let us start with the equation $$(x-y)^2\geq0,$$ which is true by the trivial inequality. Expanding, we get $$x^2-2xy+y^2\geq0.$$ Adding $4xy$ to both sides give us $$(x+y)^2\geq4xy.$$ Taking the square root of both sides and rearranging gives us

$$\dfrac{x+y}{2}\geq\sqrt{xy},$$

which is the AM-GM inequality. This inequality shows that the arithmetic mean of two numbers is always greater than the geometric mean of the two numbers, with equality when the two numbers are equal. (Exercise: show that equality holds when $x=y.$)

The AM-GM inequality can be extended to multiple variables as such: $$\dfrac{x_1+x_2+\cdots x_n}{n}\geq\sqrt[n]{x_1x_2\cdots x_n}.$$

Note that AM-GM inequality is only true for non-negative numbers. The "taking the square root" step in the proof is invalid when one (or both) of the numbers are negative.

Prove that if $a_1a_2\cdots a_n=1$ and the $a_k$ are non-negative, then $a_1+a_2+\cdots+a_n\geq n.$

Prove that, for $a, b, c > 0$,

$(abc + a)(b + c) \geq 4abc$

Prove that:

$$\frac{4}{a}+\frac{9}{b}\geq \frac{25}{a+b}$$

By Cauchy Schwartz:

$$(a+b)\left(\frac{4}{a}+\frac{9}{b}\right)\geq (3+2)^2$$

Dividing both sides by $a+b$ yields the desired inequality.

Prove that, for $x > 0$ and $m, n > 0$, that

$x^{n} + x^{m} \geq 2x^{\frac{m + n}{2}}$

Prove that for any positive real numbers $a, b, c$, $$a^4 + 4b^4 + c^2 \geq 4abc.$$

Applying the AM-GM inequality, we see that $$a^4 + 4b^4 \geq 2\sqrt{a^4\cdot4b^4} = 4a^2b^2.$$

That means $$a^4 + 4b^4 + c^2 \geq 4a^2b^2 + c^2.$$ Applying it again gives $$4a^2b^2 + c^2 \geq 2\sqrt{4a^2b^2\cdot c^2} = 4abc,$$ and we're done.

So AM GM inequality is used to prove many stuffs especially problems in Olympiads it is the basics of inequalities.

So, to recap: There's n positive numbers whose product is one, and we wanna prove their sum is at least n.

Simple! Their geometric mean is 1, since the nth root of one is one. By AM-GM, $\text{AM}\ge1$, and the sum is $\text{Sum}=\text{AM}\cdot n$, so $\text{Sum}\ge n.$