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Using just the axioms, prove the arithmetic-geometric mean inequality:

$$\sqrt{ab}\leq\frac{a+b}{2}$$

for any $a, b \in\mathbb R$ with $a > 0$ and $b > 0$. (Assume, for the moment, the existence of square roots.)

The only legal move I seem to be able to work from is the fact that since $a>0$ and $b>0$

then $a+b\gt0$

$a+b-b\gt-b\\ a\gt-b\\ a.b\gt-b.b$

Christopher Carl Heckman
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Sadio
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  • Usually from $a>0,b>0$ follows both sum and product >0, i.e. $a+b>0$ AND $ab>0.$ Do you have the product closure $ab>0$ available to use? [In short, how is $<$ defined in your text/axioms? – coffeemath Apr 05 '16 at 08:46

1 Answers1

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HINT: $\big(\sqrt{a}-\sqrt{b}\big)^2\geq0$.

EDIT: Supposing that your comment is asking "Where the hint comes from?"

We work backwards, from what we want: $$ \begin{aligned} \sqrt{ab}\leq \frac{a+b}{2}&\implies 2\sqrt{ab}\leq a+b\\ &\implies 0\leq a+b-2\sqrt{a}\sqrt{b}\\ &=a-2\sqrt{a}\sqrt{b}+b\\ &=\sqrt{a}^2-2\sqrt{a}\sqrt{b}+\sqrt{b}^2 \end{aligned}. $$ From here we must simply recognise this as a difference of two squares.

user310540
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  • thanks for the hint. I can see the path to completing the proof from the hint you've given, but can you help me see how I get to the statement in your hint? – Sadio Apr 05 '16 at 07:57
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    Do you mean "How would you have known to use $(\sqrt a-\sqrt b)^2\geq 0$?" – DanielWainfleet Apr 05 '16 at 08:05