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For all positive real values $a, b, c$, prove that
$$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{d}+\frac{d^{2}}{a} \geq a+b+c+d .$$
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Let's consider that
\begin{aligned} \frac{a^{2}}{b}+\frac{a^{2}}{b}+\frac{b^{2}}{c}+c & \geq 4 \sqrt[4]{\frac{a^{2}}{b} \frac{a^{2}}{b} \frac{b^{2}}{c} c} \\ &=4 a \\ 2 \frac{a^{2}}{b}+\frac{b^{2}}{c}+c & \geq 4 a \end{aligned}
Do it for the others to obtain
$$\begin{array}{l} 2 \frac{b^{2}}{c}+\frac{c^{2}}{d}+d \geq 4 b \\ 2 \frac{c^{2}}{d}+\frac{d^{2}}{a}+a \geq 4 c \\ 2 \frac{d^{2}}{a}+\frac{a^{2}}{b}+b \geq 4 d \end{array}$$
Adding all of them, we will get
$$3\left(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{d}+\frac{d^{2}}{a}\right)+(a+b+c+d) \geq 4(a+b+c+d)$$
$$\begin{array}{r} 3\left(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{d}+\frac{d^{2}}{a}\right) \geq 3(a+b+c+d) \\ \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{d}+\frac{d^{2}}{a} \geq a+b+c+d . \square \end{array}$$
by Diamond (75,914 points)

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