# If $a,b\ge 0$ and $a,b \in \mathbb{R}$, then prove $\frac{a+b}{2}\ge\sqrt{ab}$ [duplicate]

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, ShaileshAug 24 at 2:44

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

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

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

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

$$\sqrt{ab}\le\frac{a+b}2\iff(\sqrt{ab})^2\le\left(\frac{a+b}2\right)^2\tag2$$

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 $$a,b\in[0,\infty)$$.

HINT

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

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

therefore

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