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

Question

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?!

Answer 1

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

Answer 2

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 $$a+b\geq 2\sqrt{ab}$$

Answer 3

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

Answer 4

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

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