# Suppose that 0 < a < b. Prove that $a < \sqrt{ab} < b$ and $\sqrt{ab} \leq \frac{1}{2}(a+b)$

For part 1, I have used the NOT operator on it, giving me $a \geq \sqrt{ab} \geq b$, and then tried to prove a contradiction to the assumption. I came up with $a = b$ by transitivity, which contradicts our assumption. I'm not sure if this is right.

For part 2, I have tried proving it directly through algebra, and it didn't work, so I'm fairly certain a different proof method must be used: either contrapositive or contradiction. But that's basically as far as I can get. I can't seem to make anything logical happen via algebra.

• Yes, if you assume the negation and then derive a=b then this contradicts a<b and therefore you have proved the proposition. – Addem Mar 9 '14 at 21:56
• Thank you , the second one really has me puzzled though. – Achilles Mar 9 '14 at 21:58
• Part 1 can be done directly: $a<b$, therefore $a^2<ab$, therefore $\sqrt{a^2}<\sqrt{ab}$, and similarly for the other side. – Henning Makholm Mar 9 '14 at 22:02
• The second part is the arithmetic mean-geometric mean inequality. Note that $0 \leq (a - b)^2$, so expand the right-hand side to get the desired result. – glebovg Mar 9 '14 at 22:22
• @Addem For part 1, the negation of $\;a < \sqrt{ab} < b\;$ is $\;a \ge \sqrt{ab} \;\color{red}{\lor}\; \sqrt{ab} \ge b\;$, and from that you cannot use transitivity to conclude $\;a \ge b\;$, so there is no contradiction with $\;a \lt b\;$... – Marnix Klooster Jun 2 '17 at 22:20

For the second part, this expression is equivalent to $4ab \leq a^2 + 2ab + b^2$ since all values are positive, and this is equivalent to $0 \leq a^2 -2ab + b^2 = (a-b)^2$ which is again equivalent to $0 \leq b-a$ since $b>a$ and is equivalent to $a\leq b$.

So apparently this should be a strict inequality.

Edit: This has some circular logic at the end. To correct it, instead do this:

Suppose $0<a<b$. Then you get $0<b−a$ so $0<(b−a)^2=(a−b)^2=a^2−2ab+b^2$ which implies $4ab<a^2+2ab+b^2=(a+b)^2$ which implies $2\sqrt{ab}<a+b$ which implies $\sqrt{ab}<(a+b)/2$.

• Im not following the last part how did you conclude b > a? – Achilles Mar 9 '14 at 22:13
• I got it now thank you : ) – Achilles Mar 9 '14 at 22:14
• Ah, actually, I'm now catching a bit of circularity in my logic at that point. $0\leq (a-b)^2$ is equivalent to: either $0\leq a-b$ or $0\leq b-a$ (or, it is equivalent to $0\leq |a-b|$). But you should be able to just reverse these steps on the premise that $0<a<b$. Then you get $0<b-a$ so $0<(b-a)^2 = (a-b)^2 = a^2-2ab+b^2$ which implies $4ab<a^2+2ab+b^2 = (a+b)^2$ which implies $2\sqrt{ab} < a+b$ which implies $\sqrt{ab} < (a+b)/2$. – Addem Mar 9 '14 at 22:19
• I don't see how 0 <= (a - b)^2 implies those two inequalities ? I believe I do understand that one comes from our assumption but I do not get the other one. – Achilles Mar 9 '14 at 22:24
• Well, suppose $a=8,b=4$ then $0\leq a-b$. On the other hand, suppose $b=8,a=4$ then $0\leq b-a$. So just because you know $0\leq (a-b)^2$ does not necessarily imply $0\leq a-b$ because you could be off by a factor of -1. – Addem Mar 9 '14 at 22:24

We seek to prove $a<\sqrt{ab}<\frac{a+b}{2}<b$

We take the inequalities one at a time

Firstly, since $a<b, \sqrt{ab}>\sqrt{aa}=a$

Note that when expanding $(\sqrt{a}-\sqrt {b})^2$, which is clearly greater than zero, we obtain $a+b-2\sqrt {a}\sqrt{b}>0$, thus $a+b>2\sqrt{a}\sqrt{b}$ and $\frac{a+b}{2}>\sqrt{ab}$

Finally, since $a<b, \sqrt{ab}<\sqrt{bb}=b$

Combining these, we obtain the desired inequalities

I would start with what you need to prove, and try to simplify it using the information you have.% \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\Ref}[1]{\text{(#1)}} \newcommand{\then}{\Rightarrow} \newcommand{\when}{\Leftarrow} \newcommand{\true}{\text{true}} \newcommand{\false}{\text{false}} %

For part 1, we could calculate like this: $$\calc \tag{1} a < \sqrt{ab} < b \op\equiv\hints{square both sides of both inequalities,}\hints{no sign changes since all are positive}\hint{-- to get rid of the square root} a^2 < ab < b^2 \op\equiv\hints{divide left inequality by positive \;a\;;}\hints{divide right inequality by positive \;b\;;}\hint{-- this is the most direct simplification} a < b \;\land\; a < b \op\equiv\hint{simplify} a < b \endcalc$$ So the complex $\Ref{1}$ turns out to be equivalent to part of the assumption.

And for part 2: $$\calc \tag{2} \sqrt{ab} \leq \frac{1}{2}(a+b) \op\equiv\hints{double then square both sides of the inequality,}\hints{no sign change since both are positive}\hints{since \;a\; and \;b\; are positive}\hint{-- to get rid of the fraction and square root} 4ab \leq a^2+2ab+b^2 \op\equiv\hint{simplify} a^2-2ab+b^2 \geq 0 \op\equiv\hint{simplify -- the simplest thing to do with this} (a-b)^2 \geq 0 \op\equiv\hint{always \;x^2 \geq 0\;} \true \endcalc$$

And note that for this second part we did not need the assumption that $\;a<b\;$.