<s>[INST] Prove $\frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$. 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
 [/INST] 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 $p,q\in \{x,y,z\}$.
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