A Problem Involving an Inequality

Question

How to prove that $\frac1{a^2} + \frac1{b^2} + \frac1{c^2} \geq \frac1{ab} + \frac1{bc} + \frac1{ac}$

Assume that given symbolic terms are REAL and POSITIVE

Answer 1

moving all Terms to the left and cancelling the denominators your inequality is equivalent to $$a^2b^2+a^2c^2+b^2c^2\geq a^2bc+ab^2c+abc^2$$ this is true, since we have $$x^2+y^2+z^2\geq xy+yz+zx$$ for all real numbers

Answer 2

$$\sum_{cyc}\left(\frac{1}{a^2}-\frac{1}{ab}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)=\sum_{cyc}\left(\frac{1}{a}-\frac{1}{b}\right)^2\geq0$$

Answer 3

Note this also follows from the Cauchy-Schwarz inequality:

\begin{align} \left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)^2 & = \left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right) \left(\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{a^2}\right) \\ & \ge \left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)^2. \end{align}

Then take the square root of each side.