Prove $\frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$.

Question

So I have to prove $$ \frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.$$

I rearranged it

$$ a^2bc + ab^2c + abc^2 \leq b^2c^2 + a^2c^2 + a^2b^2 .$$

My idea from there is somehow using the AM-GM inequality. Not sure how though. Any ideas?

Thanks

Answer 1

Note that: $$a^2c^2 + a^2b^2 \ge 2a^2bc \quad\text{ by AM-GM}$$ Now add all the cyclic inequalities and you'll get the wanted inequality.

Answer 2

Hint:

Set $1/a=x$ etc.

and use $(x-y)^2\ge0$ for real $x,y$

Answer 3

Write $x=1/a$, $y=1/b$ and $z=1/c$. We get $$xy+yz+zx\leq x^2+y^2+z^2$$ Now use that $$p^2+q^2\geq 2pq$$ for each pair $p,q\in \{x,y,z\}$.

Answer 4

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