# Proof that (a^2)/(b) + (b^2)/(c) + (c^2)/(a) >= a+b+c

Proof that $\frac{{a}^{2}}{b}+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{a}\ge a+b+c$
I've tried getting everything on the left side and transforming it into something squared so that I can prove it's bigger or equals to 0 but I've been unsuccessful.
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Cullen Kelly
For positives $a$, $b$ and $c$ by C-S we obtain:
$\sum _{cyc}\frac{{a}^{2}}{b}\ge \frac{{\left(\sum _{cyc}a\right)}^{2}}{\sum _{cyc}b}=\frac{\left(a+b+c{\right)}^{2}}{a+b+c}=a+b+c.$