Given $a,\ b,\ c,\ d>0$, prove that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a} ≥ a + b + c + d.$$ I got this question from Basics of Olympiad Inequalities, Samin Riasat, and it's supposed to use the inequality of arithmetic and geometric means (AM ≥ GM), but I can't figure how.
Could you guys please help?
Thanks in advance.
Greetings from your Brazilian fellow.
Note that this is an application of $\text{AM} \ge \text{GM}$, which says that $$a^2+b^2 \ge 2ab$$
So we have $$a^2+b^2 \ge 2ab \iff \frac{a^2}{b} +b \ge 2a$$ From this, $$\sum_{cyc} \frac{a^2}{b}+\sum_{cyc} b=\sum_{cyc}\left(\frac{a^2}{b} +b\right) \ge \sum_{cyc} 2a =2\sum_{cyc} b\iff \sum_{cyc} \frac{a^2}{b} \ge \sum_{cyc} a$$ as desired.
Use Cauchy-Schwarz $$\sum_{cyc}\dfrac{a^2}{b}\sum_{cyc}b\ge (\sum_{cyc}a)^2$$ then we have $$\sum_{cyc}\dfrac{a^2}{b}\ge\sum_{cyc}a$$
Use Rearrangement $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}≥\frac{a^2}{a}+\frac{b^2}{b}+\frac{c^2}{c}+\frac{d^2}{d}=a+b+c+d$$