Show that $ab + bc + ac \ge a\sqrt{bc} + b\sqrt{ac} + c\sqrt{ab}$

Question

Where $a,b$ and $c$ are positive real numbers.

So far I have shown that $$a^2+b^2+c^2 \ge ab+bc+ac$$ and that $$a^2+b^2+c^2 \ge a\sqrt{bc} + b\sqrt{ac} + c\sqrt{ab}$$ but I am at a loss what to do next... I have tried adding various forms of the two inequalities but always end up with something extra on the side of $ab+bc+ac$. Any help appreciated!

Answer 1

Use that $$\frac{ab+ac}{2}\geq a\sqrt{bc}$$ $$\frac{ab+bc}{2}\geq b\sqrt{ac}$$ $$\frac{ac+bc}{2}\geq c\sqrt{bc}$$

Answer 2

Since $(a-b)^2\ge 0,$

we have

$$a^2+b^2\ge 2ab$$ $$b^2+c^2\ge 2bc $$ $$c^2+a^2\ge 2ca .$$

the sum gives

$$2 (a^2+b^2+c^2)\ge 2 (ab+bc+ac) $$

Answer 3

Another answer is using the "friendly" Cauchy-Schwarz inequality:

$a\sqrt{bc}+b\sqrt{ca}+ c\sqrt{ab}= \sqrt{ab}\sqrt{ac}+\sqrt{bc}\sqrt{ab}+\sqrt{ac}\sqrt{bc}\le \sqrt{(\sqrt{ab})^2+(\sqrt{bc})^2+(\sqrt{ac})^2}\times\sqrt{(\sqrt{ac})^2+(\sqrt{ab})^2+(\sqrt{bc})^2}= \sqrt{ab+bc+ca}\times\sqrt{ab+ac+bc}= \sqrt{(ab+bc+ca)^2}= ab+bc=ca$

Point is: if it can be solved by AM-GM inequality, then it can be solved by Cauchy-Schwarz inequality. These famous inequalities are like "therapeutic doctors" at a clinic...