Proving \frac{a+b}{2}\geq\sqrt{ab} for 0 < a \leq b

[Government$]

Homework Statement

Show that $\frac{a+b}{2}\geq\sqrt{ab}$ for $0 < a \leq b$

The Attempt at a Solution

Since $b \geq a$ then $b + a \geq 2a$ and $2b\geq a + b$
That is $\frac{a+b}{2}\geq a$ and $\frac{a+b}{2}\leq b$.

Since $\frac{a+b}{2}\leq b$ can i multiply $\frac{a+b}{2}\geq a$ with $b$?

If i can do that then $(\frac{a+b}{2})^{2}\geq ab$ that is:
$\frac{a+b}{2}\geq \sqrt{ab}$


On a side note if problem says prove something for all positive intigers a,b and c.
That means that a>0 , b>0, c>0 but can i take from that, $a\geq b\geq c > 0$?

 

[EulerRules]

How did you go from $b\frac{a+b}{2}\geq ab$to $(\frac{a+b}{2})^{2} \geq ab$?
You could just square both the right and left side from the start, so that
$0.25a^{2}+0.5ab+0.25b^{2} \geq ab \leftrightarrow 0.25a^{2}+0.25b^{2} \geq 0.5 ab \leftrightarrow a^{2}+b^{2}-2ab \geq 0 \leftrightarrow (a-b)^{2} \geq$0 which obviously is true.

 

[EulerRules]

As for your last question, that does not tell you very much. It tels you that a is larger than or equal to b which is larger than or equal to c which is larger than 0. A more concise way to express the stipulation is to write
$a,b,c \in N$
or
$a,b,c \in Z^{+}$

 

[Government$]

Well i thought that since b≥ (a+b)/2 and (a+b)/2 * b = ((a+b)/2)^2 (because it can be b = (a+b)/2)

 

[Mark44]

Well i thought that since b≥ (a+b)/2 and (a+b)/2 * b = ((a+b)/2)^2 (because it can be b = (a+b)/2)

But you can't treat this as an equality, which is what you seem to be doing.

 

[Curious3141]

Since $\frac{a+b}{2}\leq b$ can i multiply $\frac{a+b}{2}\geq a$ with $b$?

If i can do that then $(\frac{a+b}{2})^{2}\geq ab$ that is:
$\frac{a+b}{2}\geq \sqrt{ab}$

You can't do this: the inequality signs face different ways.

If $a \geq b$ and $c \geq d$, then for positive a,b,c and d, you CAN assert that $ac \geq bd$. If that's a $\leq$ sign, then the opposite applies (just switch all the signs). However, if the signs are different you cannot conclude anything from multiplying the two inequalities.

The simplest way to approach this is to start by observing that $(\sqrt{a} - \sqrt{b})^2 \geq 0$ for all $a,b \in \mathbb{R^+}$ (positive real a and b).

On a side note if problem says prove something for all positive intigers a,b and c.
That means that a>0 , b>0, c>0 but can i take from that, $a\geq b\geq c > 0$?

No you cannot infer that. You can only say that a, b and c are all greater than 0, but you cannot conclude anything about how they're ordered with respect to one another.

However, if a, b and c are *arbitrary* positive integers, and the property you're required to prove is essentially symmetric in a, b, and c (i.e. switching the symbols around doesn't matter), then you can start by arguing "without loss of generality, let $a \geq b \geq c > 0$" if it helps your proof. This is a common technique.