Inequalities Proof: if 0<=a<=b, then a<=sqrt(ab)<=(a+b)/2<=b

[MuchJokes]

I have slight issues with proofs, and have no idea how to even START this one, so I'd like a hand up if possible.

I also have no idea if this is the right section, but it looked right.

Given that 0 <= a <= b show that

a <= sqrt(ab) <= (a + b)/2 <= b

I just don't know what the style is for these kind of proofs, I'm used to left side, right side kind of things, and this is pretty new to me.

Thank you, and if someone could tell me if there's a program installed on the forum to write real math symbols, that would be great.

 

[pka]

It is always that the average of two numbers is between them.
\(\displaystyle a < b \Rightarrow \quad \frac{a}{2} < \frac{b}{2} \Rightarrow \quad \frac{a}{2} + \frac{a}{2} < \frac{b}{2} + \frac{a}{2} \Rightarrow \quad a < \frac{{b + a}}{2}\).
You can complete to get \(\displaystyle a < \frac{{b + a}}{2} < b\) by adding \(\displaystyle \frac {b}{2}\).

Now this is true: \(\displaystyle 0 \leqslant \left( {x - y} \right)^2 = x^2 - 2xy + y^2 \Rightarrow \quad 2xy \leqslant x^2 + y^2 \Rightarrow \quad xy \leqslant \frac{{x^2 + y^2 }}{2}\).

Because each of a & b is positive: \(\displaystyle x = \sqrt a \,\& \,y = \sqrt b \Rightarrow \quad \sqrt {ab} \leqslant \frac{{a + b}}{2}\)

You can put all of that together?