AM-GM inequality: $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq a + b + c + d$

Question

Let $a, b, c, d > 0$. Prove that $$\dfrac{a^2}{b} + \dfrac{b^2}{c} + \dfrac{c^2}{d} + \dfrac{d^2}{a} \geq a + b + c + d.$$

I'm supposed to prove this by $AM$-$GM$, but I can't see how. Any help would be appreciated.

Answer 1

Using this inequality is probably the quickest way $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\geq \frac{(a+b+c+d)^2}{b+c+d+a}=a+b+c+d$$

Otherwise, using AM-GM we have: $$\frac{a^2}{b}+b+\frac{b^2}{c}+c+\frac{c^2}{d}+d+\frac{d^2}{a}+a \geq 2\sqrt{\frac{a^2}{b}b}+2\sqrt{\frac{b^2}{c}c}+2\sqrt{\frac{c^2}{d}d}+2\sqrt{\frac{d^2}{a}a}=2(a+b+c+d)$$

Answer 2

By AM-GM $\frac{a^2}{b}+b\geq2a$.

Thus, $\sum\limits_{cyc}\frac{a^2}{b}\geq\sum\limits_{cyc}(2a-b)=\sum\limits_{cyc}a$ and we are done!

Answer 3

Tl;dr: The proposition and it's more obvious generalisation (for more or less than four terms also) follows trivially from the left-hand side of the rearrangement inequality.

Explanation: The left side of the rearrangement inequality tells us that

$$ x_{n}y_{1}+\cdots +x_{1}y_{n}\leq x_{\sigma (1)}y_{1}+\cdots +x_{\sigma (n)}y_{n }$$ for every choice of real numbers

$$x_{1}\leq \cdots \leq x_{n}\quad {\text{ and }}\quad y_{1}\leq \cdots \leq y_{n}.$$

In particular, let's say we are given positive real numbers $\ (a_k)_{1\leq k\leq n},\ $ and are asked to prove that $\ \frac{ {a_1}^2}{ a_2 } + \frac{ {a_2}^2}{ a_3 } + \ldots + \frac{ {a_{n-1}}^2}{ a_n } + \frac{ {a_n}^2}{ a_1 } \geq a_1 + \ldots + a_n,\ $ which is the desired generalisation of the original question.

Re-arrange the order of terms in the sum $\ S = \frac{ {a_1}^2}{ a_2 } + \frac{ {a_2}^2}{ a_3 } + \ldots + \frac{ {a_{n-1}}^2}{ a_n } + \frac{ {a_n}^2}{ a_1 }\ $ to this: $\ S = \frac{ {a_{i_1} }^2}{a_{j_1}} + \frac{ { a_{i_2} }^2 }{a_{j_2}} + \ldots + \frac{ { a_{i_{n-1}} }^2 }{a_{j_{n-1}}} + \frac{ { a_{i_n} }^2 }{a_{j_n}},\ $ so that $\ \frac{1}{a_{j_1} } \leq \frac{1}{a_{j_2} } \leq \ldots \leq \frac{1}{a_{j_{n-1}} } \leq \frac{1}{a_{j_n} }.\ $ These $\ \frac{1}{a_{j_k}}\ $ are the "$\ y_k\ $" in the rearrangement inequality, that is: $\ y_1:= \frac{1}{a_{j_1} },\ y_2:= \frac{1}{a_{j_2} },\ $ and so on. The $\ {a_{i_k} }^2\ $ are the " $ x_k\ $ " in the rearrangement inequality, that is: $\ x_1 = {a_{i_1} }^2,\ x_1 = {a_{i_2} }^2,\ $ and so on...

Since all the terms $\ y_1,\ y_2,\ \ldots,\ $ are positive, it follows that $\ a_{j_1} \geq a_{j_2} \geq \ldots \geq a_{j_{n-1}} \geq a_{j_n}.\ $ If $\ y_1,\ y_2,\ \ldots,\ $ were allowed to be negative then this would not be the case. The rearrangement inequality above tells us the minimum possible value of $\ S\ $ is $\ \frac{ {{a_{j_1}} ^2 } }{a_{j_1}} + \frac{ { {a_{j_2} }^2 } }{a_{j_2}} + \ldots + \frac{ {a_{j_{n-1}} }^2 }{a_{j_{n-1}}} + \frac{ { a_{j_n} }^2 } {a_{j_n}}\ = a_{j_1} + a_{j_2} + \ldots + a_{j_{n-1}} + a_{j_n}.\ $ In other words, every possible permutation of $\ (a_{i_k})_{1\leq k\leq n}\ $ for $\ S\ $ is $\ \geq a_{j_1} + a_{j_2} + \ldots + a_{j_{n-1}} + a_{j_n},\ $ and we are done.