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We have two positive numbers $a$ and $b$. It surely means that they might be $0$ or bigger than $0$ (and they are real numbers). $$a,b\ge 0$$ $$a,b \in \mathbb{R}$$ So now we must prove that: $$\frac{a+b}{2}\ge\sqrt{ab}$$ In fact I don't know how to use relations of fractions to prove it. Would you explain an easy way?!


marked as duplicate by Feng Shao, The Count, Shogun, nmasanta, Shailesh Aug 24 at 2:44

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  • $\begingroup$ Should it not be $$\frac{a+b}{2}\geq \sqrt{ab}$$? $\endgroup$ – Dr. Sonnhard Graubner Nov 3 '18 at 19:27
  • $\begingroup$ Yes Dr. I have edit it $\endgroup$ – user602338 Nov 3 '18 at 19:28
  • 1
    $\begingroup$ Do you have any ideas?! $\endgroup$ – user602338 Nov 3 '18 at 19:28
  • $\begingroup$ It's called A.M( Arithmetic mean)$\geq$ G.M( Geometric mean) $\endgroup$ – Surajit Nov 3 '18 at 19:30
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    $\begingroup$ This must be a duplicate $\endgroup$ – Alfred Yerger Nov 3 '18 at 20:19

Hint: $$a^2+b^2\geq 2ab$$ or $$(a-b)^2\geq 0$$ Expanding gives $$a^2+b^2-2ab\geq 0$$ and $$a^2+b^2+2ab\geq 4ab$$ so $$(a+b)^2\geq 4ab$$ taking the square root we get $$a+b\geq 2\sqrt{ab}$$

  • $\begingroup$ So would you tell me a path in addition to this hint doctor? $\endgroup$ – user602338 Nov 3 '18 at 19:31
  • $\begingroup$ Thanks a lot Dr. I WILL ACCEPT ALL OF THE ANSWERS ! $\endgroup$ – user602338 Nov 3 '18 at 19:55
  • $\begingroup$ Hh impossible to do that but thanks a lottt! $\endgroup$ – user602338 Nov 3 '18 at 19:56
  • $\begingroup$ Or, after your first line, replace $a$ and $b$ by $\sqrt a$ and $\sqrt b$. $\endgroup$ – Lord Shark the Unknown Nov 3 '18 at 20:13

Hint: $$(\sqrt{a}-\sqrt{b})^2 \geq 0$$

  • $\begingroup$ I think that it is an acceptable answer! In addition to what Dr.sonnhard says but it is easier! $\endgroup$ – user602338 Nov 3 '18 at 19:37
  • $\begingroup$ @user602338 Thank you! $\endgroup$ – Botond Nov 3 '18 at 19:42

The key point to prove it is knowing that strictly increasing functions preserves inequalities in the real line (by the definition of being strictly increasing), that is, suppose that $f:X\to\Bbb R$ is a strictly increasing function (for some $X\subset\Bbb R$), then

$$r\le s\iff f(r)\le f(s)\tag1$$

for any pair $r,s\in X$.

In your case the function $f:[0,\infty)\to\Bbb R,\, x\mapsto x^2$ is strictly increasing, and because $a,b\in[0,\infty)$ you have that


Then rearranging the RHS on $(2)$ you find that

$$\begin{align}(\sqrt{ab})^2\le\left(\frac{a+b}2\right)^2&\iff 4ab\le a^2+2ab+b^2\\&\iff0\le a^2-2ab+b^2=(a-b)^2\end{align}\tag3$$

And because $0\le(a-b)^2$ is clearly true then we conclude that the original inequality $\sqrt{ab}\le\frac{a+b}2$ also holds for all pairs $a,b\in[0,\infty)$.



Recall that since $f(x)$ is an increasing function for $x\ge 0$ we have

$$A\ge B \iff A^2\ge B^2 \quad A,B\ge 0$$


$$\frac{a+b}{2}\ge\sqrt{ab}\iff \left(\frac{a+b}{2}\right)^2\ge \left(\sqrt{ab}\right)^2$$


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