How do I show that if $a_1\neq a_2$, then $\left (\frac{a_1+a_2}{2} \right )^{2}> a_1a_2$.

Question

My work: $(a_1-a_2)^{2}> 0\Rightarrow a_1^2-2a_1a_2 +a_2^2>0\Rightarrow a_1^2+a_2^2> 2a_1a_2\Rightarrow \frac{a_1^2+a_2^2}{2}> a_1a_2$.

Why isn't this working?

The argument is correct. The second statement in the chain should have a $> 0$ at the end. — Utku Alhan, Mar 22 '14 at 21:59
@UtkuAlhan Can't you see that I am getting something different? — OGC, Mar 22 '14 at 22:00
They are stumped on how to go from $\frac{a_1^2+a_2^2}{2} > a_1a_2$ to $\left( \frac{a_1+a_2}{2} \right)^2 > a_1a_2$ — ruler501, Mar 22 '14 at 22:02
You should go $$a_1^2 -2a_1a_2 + a_2^2 > 0 \Rightarrow a_1^2 + 2a_1 a_2 + a_2^2 > 4 a_1 a_2.$$ — Daniel Fischer, Mar 22 '14 at 22:03

score 6 · Accepted Answer · answered Mar 22 '14 at 22:03

6

You're almost there. If we add $a_1a_2$ to the final inequality we get $$ \frac{a_1^2 + a_2^2 + 2a_1a_2}{2} > 2a_1a_2. $$ Now factor the left side and divide by $2$ to finish the problem.

answered Mar 22 '14 at 22:03

user134824

12,212

If you divide both sides by two you don't get the desired form(it would be over 4 on the left wouldn't it). – ruler501 Mar 22 '14 at 22:06
1

@ruler501 Exactly! Since $\sqrt{4}=(\pm)2$, this works! – Mar 22 '14 at 22:09
ah. I forgot you had to move the $\frac{1}{4}$ into the parentheses. – ruler501 Mar 22 '14 at 22:09

score 3 · Answer 2 · answered Mar 22 '14 at 22:11

3

Ok, just another way of thinking:

By absurd, suppose that $$\left( \frac{a_1+a_2}{2}\right)^2\leq a_1\cdot a_2. $$ Then, $$ a_1^2+2a_1a_2+a_2^2\leq 4a_1a_2\Rightarrow a_1^2-2a_1a_2+a_2^2\leq 0$$ $$\Rightarrow (a_1-a_2)^2\leq 0. $$ As the square of a real number can not be negative, we obtain $$a_1-a_2=0, $$ i.e., $$a_1=a_2, $$ but this is absurd, since by hypothesis $a_1\neq a_2$.

Thus, $$\left( \frac{a_1+a_2}{2}\right)^2\geq a_1\cdot a_2. $$

answered Mar 22 '14 at 22:11

ZHN

1,419

1

I am not seeing the point/gain here - this is essentially a direct proof disguised as a fancy contradiction. – ocg Mar 22 '14 at 22:26
It was just another way of thinking, that's all. – ZHN Mar 22 '14 at 22:32

score 1 · Answer 3 · answered Mar 22 '14 at 22:19

1

In general you got this inequality $a^{\lambda}b^{\lambda-1 } \le \lambda a+(1-\lambda)b$ .(not hard to prove either)

answered Mar 22 '14 at 22:19

Eli Elizirov

1,820

score 0 · Answer 4 · answered Mar 22 '14 at 22:06

You basically are trying to prove the AM-GM inequality, for $n=2$.

Let me show this: $$\left(\dfrac{a_1+a_2}{2}\right)^2-a_1a_2=\dfrac{1}{4}(a_1^2+a_2^2+2a_2a_1)-a_2a_1\\ =\dfrac{1}{4}(a_1^2+a_2^2-2a_1a_2)=\left(\dfrac{a_1-a_2}{2}\right)^2>0$$ This proves the desired statement.

score 0 · Answer 5 · answered Mar 22 '14 at 22:11

Others have commented on the algebraic proof that the statement is true. This is too short for a comment; here I will give a geometric explanation for the statement:

First of all, you probably know that a square has a larger area than a rectangle if they have equal perimeters. The proof of this statement is exactly what other answers have given. Let the two sides of a rectangle be $a_1$ and $a_2$. Then the square with equal perimeter to this will have side lengths of $\frac{a_1 + a_2}{2}$. This square will have a greater area than the rectangle unless the rectangle is a square, too, but since $a_1 \neq a_2$ we cannot have that. Thus, $\left( \frac{a_1 + a_2}{2} \right)^2 > a_1 a_2$

How do I show that if $a_1\neq a_2$, then $\left (\frac{a_1+a_2}{2} \right )^{2}> a_1a_2$.

5 Answers5