Prove that, $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\ge 4$ where we do not use AM-GM inequality on the given statement to prove it.

Prove that, $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}\ge 4$ where we do not use AM-GM inequality on the given statement to prove it. Typically, I am actually looking for a little advanced and elegant solution.

EDIT: $a,b,c,d>0$

Solution 1:

Not exactly AM-GM, but Rearrangement Inequality immediately gives

$$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \ge \frac{a}{a}+\frac{b}{b}+\frac{c}{c}+\frac{d}{d} = 4$$

Solution 2:

For $N > 0$, consider the set $$ D_N = \left\{ (a,b,c,d) \in \mathbb R^4 \, | \, \frac 1N \le a,b,c,d \le N \right\}. $$ This set is compact, and the function $f(a,b,c,d) \overset{def}= \frac ab + \frac bc + \frac cd + \frac da$ is continuous on $D_N$, hence we have the existence of a minimum. Without loss of generality, we can assume that there is a minimizer in the interior of $D_N$, since if it is on the boundary for some $N$, it is in the interior of a $D_N$ for $N$ slightly bigger. (Note that when any component approaches $0$ or $\infty$, $f \to \infty$, hence the minimum is not there, which allows us to consider the interior only and dismiss boundaries.)

Computing the gradient, one gets $$ \nabla f(a,b,c,d) = \left( \frac 1b - \frac d{a^2}, \frac 1c - \frac a{b^2}, \frac 1d - \frac b{c^2}, \frac 1a - \frac c{d^2}\right) $$ and since our minimizer will be in the interior, there exists a minimum (A,B,C,D) such that $$ A^2 = BD, B^2 = CA, C^2 = DB, D^2 = AC \implies A=B=C=D \implies f(A,B,C,D) = 4. $$ Since $4$ is the minimum value for all $N$, it is the minimal value.

Hope that helps,

Solution 3:

Let's prove that $\frac{a}{a'} + \frac{b}{b'} + \frac{c}{c'} + \frac{d}{d'} \geq 4$ if $(a', b', c', d')$ is the tuple $(a, b, c, d)$ shuffled in an arbitrary order. Your inequality will follow as a special case when $(a', b', c', d')=(b, c, d, a)$.

When the problem is formulated like this, we can safely assume that $a \leq b \leq c \leq d$. Let's do that to make things easier.

If $(a', b', c', d') \neq (a, b, c, d)$, then there's a pair of adjacent elements in the tuple $(a', b', c', d')$ that goes in decreasing order. For instance, let's assume that $a' > b'$. Form a new tuple $(a'', b'', c'', d'') = (b', a', c', d')$. We have swapped positions of the two elements so that now they go in the ascending order. Observe that since $a' > b'$ and $a < b$, we have $$ \frac{a}{a''} + \frac{b}{b''} = \frac{a}{b'} + \frac{b}{a'} \leq \frac{a}{a'} + \frac{b}{b'}, $$ therefore $$ \frac{a}{a''} + \frac{b}{b''} + \frac{c}{c''} + \frac{d}{d''} \leq \frac{a}{a'} + \frac{b}{b'} + \frac{c}{c'} + \frac{d}{d'}. $$

Using this observation, we can do the following. We can take the tuple $(a', b', c', d')$ and bubble sort it. Each time when we swap places of two adjacent elements that go in descending order, the value $\frac{a}{a'} + \frac{b}{b'} + \frac{c}{c'} + \frac{d}{d'}$ decreases, as we have just shown. When we are done with the bubble sort, we will have the tuple that goes in ascending order, i.e. $(a', b', c', d') = (a, b, c, d)$. For this tuple we have $\frac{a}{a} + \frac{b}{b} + \frac{c}{c} + \frac{d}{d} = 4$. Since the value was only decreasing during the bubble sort, the original value was greater or equal than the final one. Therefore $\frac{a}{a'} + \frac{b}{b'} + \frac{c}{c'} + \frac{d}{d'} \geq 4$, for an arbitrary arrangement $(a', b', c', d')$, qed.