Suppose that $a,b \in \mathbb{R}$ are positive.

Prove that:

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

Note: This inequality is known as the inequality between arithmetic mean, $\frac{a + b}{2}$, and geometric mean, $\sqrt{ab}$, in short, the am-gm inequality.

Hint: Use the fact that $(x,y)^2 \geq 0$ for all numbers $x$ and $y$.

So far we only learned how to prove by Induction.

My solution:

$\sqrt{ab} \leq \frac{a + b}{2}$ --> can be written as =

$$\frac{x_1+ \ldots + x_n}{n} \geq \sqrt[n]{x_1 \cdots x_n}$$

Base case: $n=1$

$$\frac{x_1}{1} \geq \sqrt[1]{x_1}$$ ---> True $x_1 = x_1$

Inductive step: Assume true for $n=k$

$$\frac{x_1+ \ldots + x_k}{k} \geq \sqrt[k]{x_1 \cdots x_k}$$

Show true for n=k+1

$$\frac{x_1+ \ldots + x_{k+1}}{k+1} \geq \sqrt[k+1]{x_1 \cdots x_{k+1}}$$

I am stuck at this step. Please help Thank you very much.

  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ Sep 20 '15 at 9:45
  • $\begingroup$ @N.F.Taussig Yes sir $\endgroup$
    – question
    Sep 20 '15 at 9:46
  • 1
    $\begingroup$ I also suggest to search for the arithmetic-geometric inequality and induction on this site. You will find several questions. $\endgroup$
    – mickep
    Sep 20 '15 at 9:47
  • 3
    $\begingroup$ @mickep I found one math.stackexchange.com/questions/691807/… There is one part I dont really understand: the part that i can not figure our $\endgroup$
    – question
    Sep 20 '15 at 9:52

Your theorem is only about the $n=2$ case, so I don't see the need to prove a more general statement. Just set $x=\sqrt{a}$, $y=\sqrt{b}$ and follow the hint given: $$ 0\le(x-y)^2=x^2+y^2-2xy=a+b-2\sqrt{ab}, $$ Q.E.D.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.