Sohail Farhangi Submitted 9 October, 2015

Prove the following inequality, for positive real numbers $a, b,$ and $c,$ and determine when the equality holds:

$a\sqrt{4a^2+5bc}+b\sqrt{4b^2+5ca}+c\sqrt{4c^2+5ab} \ge (a+b+c)^2$.

Solution

Observe that the above is equivalent to the inequality below:

$a\sqrt{4a^2+5bc}-a^2-ab-ac+b\sqrt{4b^2+5ca}-ab-b^2-bc+c\sqrt{4c^2+5ab}-ca-bc-c^2 \ge 0$.

Now we define the auxiliary variables so that we may restate the latter inequality as follows:

$v = \left(\begin{array}{ccc} a \\ b \\ c \end{array}\right),$ $\displaystyle A = \left( \begin{array}{ccc} \sqrt{4+5\frac{bc}{a^2}}-1 & -1 & -1 \\ -1 & \sqrt{4+5\frac{ca}{b^2}}-1 & -1 \\ -1 & -1 & \sqrt{4+5\frac{ab}{c^2}}-1 \end{array} \right)$

(1)

$v^tAv \ge 0$

If we can show that $A$ is a positive semi-definite matrix, we will only need to find the eigenvectors corresponding to the eigenvalue of $0$ to resolve the problem. To show that the Hermitian matrix $A$ is positive semi-definite, we only need to verify that all principal minors are nonnegative. It is clear that all $1\times 1$ minors of $A$ are positive. Next, if we note that $\displaystyle\sqrt{4+5\frac{bc}{a^2}}-1 \ge \sqrt{4+5\cdot 0}-1 = 1$, it is easily seen that all $2\times 2$ minors of $A$ are nonnegative. Lastly, we need to verify that the determinant of $A$ is nonnegative, and this is equivalent to (2) below.

To make the formulas more manageable let's introduce $\displaystyle U=\sqrt{4+5\frac{bc}{a^2}},$ $\displaystyle V=\sqrt{4+5\frac{ca}{b^2}},$ $\displaystyle W=\sqrt{4+5\frac{ab}{c^2}}.$ Then

\displaystyle\begin{align} det(A) &= (U-1)(V-1)(W-1) - 2(-1)(-1)(-1)-(U-1)-(V-1)-(W-1)\\ &=UVW-UV-VW-WU \end{align}

The determinate $|A|$ is not negative only if $UVW\ge UV+VW+WU.$ Since all three are greater than $0,$ this is equivalent to $\displaystyle1\ge\frac{1}{U}+\frac{1}{V}+\frac{1}{W},$ or using the original values,

$\displaystyle 1\ge (4+5\frac{ab}{c^2})^{-\frac{1}{2}}+(4+5\frac{bc}{a^2})^{-\frac{1}{2}}+(4+5\frac{ca}{b^2})^{-\frac{1}{2}}.$

Observe that the three expression are homogeneous in the variables $a,$ $b,$ $c$ such that we may assume $abc = 1.$ We may now perform the following substitution. $x = a^{-3}, y = b^{-3},$ and $z = c^{-3}$ to obtain

(2)

$1 \ge (4+5x)^{-\frac{1}{2}}+(4+5y)^{-\frac{1}{2}}+(4+5z)^{-\frac{1}{2}}$, $xyz = 1$.

The latter is a known inequality with two equality cases, so $A$ is positive semi-definite as desired. The inequality (2) will be treated on a separate page. We will now consider the two equality cases in order to find when equality holds in the original problem.

Case 1: $x = y = z = 1$. In this case it is evident that $a = b = c$, and plugging these values in $A$ yields the matrix $A_1$ below.

$A_1 = \left( \begin{array}{ccc} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{array} \right)$

It can be seen that the eigenvalues of $A_1$ are $0, 3,$ and $3$, so the eigenspace corresponding to the eigenvalue $0$ has dimension $1$. We can see that

$\left(\begin{array}{ccc} 1 \\ 1 \\ 1 \end{array}\right)$

is an eigenvector of $A_0$ corresponding to the eigenvalue of $0$, and it follows that equality in (1) holds if $a = b = c$.

Case 2: $\displaystyle\{x, y, z\} = \lim_{n \rightarrow \infty}\{\frac{1}{n}, \frac{1}{n}, n^2\}$. For shorthand, we will denote this solution as $\{x, y, z\} = \{0, 0, \infty\}$. Moreover, it can be seen that there are 3 permutations of this solution, but it is only necessary to consider 1 of them. In this case it is evident that $a = b = \infty$, and $c = 0$. We can see that plugging these values into $A$ yields the matrix $A_2$ below.

$A_2 = \left( \begin{array}{ccc} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & \infty \end{array} \right)$

It can be see that

$v_1 = \left(\begin{array}{ccc} 1 \\ 1 \\ 0 \end{array}\right)$

Is an eigenvector corresponding to the eigenvalue $0$. Furthermore, we can see that $v_1$ is the only eigenvector corresponding to the eigenvalue $0$ as follows. Consider the vector $v_2$ below.

$v_2 = \left(\begin{array}{ccc} r \\ s \\ t \end{array}\right)$

In order for $v_2$ to be an eigenvector corresponding to the eigenvalue $0$ of the matrix $A_2$, we must have $t = 0$ as seen from the fact that the entry in the third row and third column of $A_2$ is $\infty$, while the other entries in the third row are finite. Examining the first row of $A_2$ tells us that $r = s$, so we can see that $v_1$ is the only eigenvector of $A_2$ corresponding to the eigenvalue $0$ as desired, and this corresponds to the solution $a = b$, and $c = 0$ for (1).

Exercise

Consider a function of the form $f(a,b,c) = \sqrt{4+\displaystyle\sum_{i = 0}^n d_ia^{r_i}b^{s_i}c^{t_i}}$ where $\{d_i\}_{i = 0}^{\infty}, \{r_i\}_{i = 0}^{\infty}, \{s_i\}_{i = 0}^{\infty},$ and $\{t_i\}_{i = 0}^{\infty}$ are all real sequences with $d_i > 0,$ $1 \le i \le n$, and $f(1,1,1) = 3$. For positive real numbers $a, b,$ and $c$, and any real number $x, y,$ and $z$, prove the following inequality, given that $abc \ge 1$.

$x^2f(a,b,c)+y^2f(c,a,b)+z^2f(b,c,a) \ge (x+y+z)^2$.

What can be said about the equality case? The equality case does have some dependence on the function $f$.