Let a and b be real numbers such that 0 < a < b. Prove $\frac{a+b}2 > \sqrt{ab} > \frac{2ab}{a + b}$

Question

How can I prove this? Been working for hours and got nowhere. I see $\frac{a+b}{2}$ and $\frac{2ab}{a + b}$ are almost reciprocals. Is this important?

Please explain step by step.

Square the inequalities. Then use $0 <(a-b)^2 = a^2-2ab+b^2$. — marty cohen, Sep 20 '15 at 21:54

score 3 · Accepted Answer · answered Sep 20 '15 at 21:57

3

Hint:

The left-hand side inequality is equivalent to $a+b-2\sqrt{ab}$.

For the right-hand side, replace $a$ and $b$ with $\dfrac1a$ and $\dfrac1b$ in the first inequality.

answered Sep 20 '15 at 21:57

Bernard

175,478

How do I prove $a + b - 2\sqrt{ab}$ is positive? – Tom Stanton Sep 20 '15 at 22:20
Set $x=\sqrt a, y=\sqrt b$. – Bernard Sep 20 '15 at 22:24

score 0 · Answer 2 · 2015-12-29T07:59:34.437

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first we show that$ (a+b)\over 2$ > $\sqrt {ab}$

i.e. T.S. $a+b \gt2 \sqrt {ab}$

i.e. T.S. $(a+b)^2 \gt 4ab$

i.e. T.S. $a^2+b^2 \gt2ab$

i.e.T.S. $(a-b)^2 >0$

which is true.

Similarly you can prove another inequality.

edited Dec 29 '15 at 07:59

answered Sep 20 '15 at 22:01

Passionate · Answer 3 · 2017-02-08T12:57:03.610

0

There are some relations between Arithmetic Mean, Geometric Mean and Harmonic Mean.

A.M. ≥ G.M. ≥ H.M. & A.M. × H.M. = G.M.^2

Prove of A.M.≥G.M.≥H.M.

edited Feb 08 '17 at 12:57

answered Feb 06 '17 at 18:10

Passionate

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Small correction: $AM \times HM = GM^2$. – Brian Tung Feb 06 '17 at 18:18

Let a and b be real numbers such that 0 < a < b. Prove $\frac{a+b}2 > \sqrt{ab} > \frac{2ab}{a + b}$

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