How do I show that if $a_1\neq a_2$, then $\left (\frac{a_1+a_2}{2} \right )^{2}> a_1a_2$. How do I show that if $a_1\neq a_2$, then $\left (\frac{a_1+a_2}{2}  \right )^{2}> a_1a_2$.
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?
 A: 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.
A: 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,
$$\left( \frac{a_1+a_2}{2}\right)^2\geq a_1\cdot a_2. $$
A: In general you got this inequality $a^{\lambda}b^{\lambda-1 } \le \lambda a+(1-\lambda)b$ .(not hard to prove either)
A: 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.
A: 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, $\left( \frac{a_1 + a_2}{2} \right)^2 > a_1 a_2$
