Show that $\forall (a, b, c) \ \in \mathbf{R}: \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} \ge \frac{b}{a} + \frac{c}{b} + \frac{a}{c}$

Question

Show that

\begin{equation} \forall (a, b, c) \ \in \mathbf{R}: \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} \ge \frac{b}{a} + \frac{c}{b} + \frac{a}{c} \end{equation}

My approach was as follows: multiplying both sides by $a^2b^2c^2$, we get

\begin{equation} a^4c^2 + b^4a^2 + c^4b^2 \ge b^3c^2a + c^3a^2b + a^3b^2c \end{equation}

Now, introduce the following notation:

\begin{equation} \begin{bmatrix} a_1 & a_2 & \dots & a_n \\ b_1 & b_2 & \dots & b_n \end{bmatrix} = a_1b_1 + a_2b_2 + \dots + a_nb_n. \end{equation}

Thus, we can write the above inequality as:

\begin{gather*} a^4c^2 + b^4a^2 + c^4b^2 \ge b^3c^2a + c^3a^2b + a^3b^2c \\ \iff a^3c^2a + b^3a^2b + c^3b^2c \ge b^3c^2a + c^3a^2b + a^3b^2c \\ \iff \begin{bmatrix} a^3 & b^3 & c^3 \\ c^2a & a^2b & b^2c \end{bmatrix} \ge \begin{bmatrix} a^3 & b^3 & c^3 \\ b^2c & c^2a & a^2b \end{bmatrix} \end{gather*}

Now, assume that $a \le b\le c$ (note that the inequality is not symmetric, but the same argument as is about to be made could be made assuming any other order of the variables as well). If we look at the $RHS$ of the above inequality, we see that we have the two arrays $A = (a^3, b^3, c^3), B = (b^2c, c^2a, a^2b)$. If $a \le b \le c$, we have that $a^3 \le b^3 \le c^3$, which is equivalent to $A$ being strictly increasing. Moreover, we see that the last element of $B$ ($a^2b$) is the smallest element in $B$ (this is trivial to show). Now, according to the rearrangement inequality, the smallest possible sum (for $n=2$) is achieved when $A$ is increasing and $B$ is decreasing, which means that the sum in $RHS$ is either the smallest possible sum or the second smallest, depending on if $b^2c \ge c^2a$ or not. But the problem is that this very same argument about smallest or second smallest sum can be made about the $LHS$ as well...

Does anyone have any ideas on how to proceed?

Answer 1

Using AM-GM, we have $$\frac{a^2}{b^2} + \frac{b^2}{c^2} \ge 2\sqrt{\frac{a^2}{b^2} \cdot \frac{b^2}{c^2}} = 2\sqrt{\frac{a^2}{c^2}} \ge \frac{2a}{c},$$ $$\frac{b^2}{c^2} + \frac{c^2}{a^2} \ge 2\sqrt{\frac{b^2}{c^2} \cdot \frac{c^2}{a^2}} = 2\sqrt{\frac{b^2}{a^2}} \ge \frac{2b}{a},$$ $$\frac{c^2}{a^2} + \frac{a^2}{b^2} \ge 2\sqrt{\frac{c^2}{a^2} \cdot \frac{a^2}{b^2}} = 2\sqrt{\frac{c^2}{b^2}} \ge \frac{2c}{b}.$$ Add them up to get the desired inequality.

Answer 2

Here's one way to solve the problem while avoiding algebra. The key idea is that $\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a} = 1$. Thus, we will substitute $x = \frac{a}{b}, y = \frac{b}{c},$ and $z = \frac{c}{a}$. The equation that we want to show is now $$ x^2 + y^2 + z^2 \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $$ under the condition that $xyz = 1$. However, we can now multiply $xyz$ to the right hand side to show that we want $$ x^2 + y^2 + z^2 \geq yz + xz + xy $$ which is true as it rearranges to $\frac{1}{2} \left( (x-y)^2 + (x-z)^2 + (y-z)^2 \right) \geq 0$. As all of our steps are reversible, we are done.

Answer 3

Note that you can express your condition $0\le a\le b\le c$ by setting $\begin{cases}b=au\\c=auv\end{cases}\quad$ with $u,v\ge 1$.

This leads to $$\frac 1{u^2}+\frac 1{v^2}+u^2v^2\ge u+v+\frac 1{uv}$$

Then it is less straightforwad than ETS1331's way, but similarly numerator of the difference can be written as

$$(u-v)^2+uv(u^2v-1)(uv^2-1)\ge 0$$

Which is positive provided $u,v$ are greater than $1$, but we restricted to positive numbers, so the other solution is superior. I wanted to show it anyway.

Answer 4

It should be noted that the result is obviously nonsensical if any one of $\ a,b,c\ $ are equal to $\ 0,\ $ so suppose this is not the case.

Suppose $\ u,v,w \in \mathbb{R},\ $ and $\ \max\{u,v,w\} = w.\ $ Then,

$$ vu + w^2 \geq vw + wu. \qquad (1) $$

Thus, for all $\ u,v,w \in \mathbb{R}\ $ with $\ \max\{u,v,w\} = w,\ $

$$ (u-v)^2 \geq 0, \implies u^2 + v^2 \geq uv + vu,$$

$$\implies u^2 + v^2 + w^2 \geq uv +vu + w^2 \overset{(1)}{\geq} uv + vw + wu. $$

Now let $\ w = \max \{ ca^2, ab^2, bc^2 \}.\ $ Due to cyclicity, we may suppose WLOG that $\ w\ $ is any one of these: so suppose WLOG that $\ w=bc^2.\ $ Then, substituting into the above, we get:

$$ c^2 a^4 + a^2 b^4 + b^2 c^4 \geq a^3 b^2 c + b^3 c^2 a + c^3 a^2 b. $$

Since none of $\ a,b,c\ $ are equal to zero, we may divide through by $\ a^2b^2c^2\ $ and obtain the result.