# Proving this inequality for positive real numbers a, b, c, d?

## $\frac{a}{b + 2 \cdot c + 3 \cdot d} + \frac{b}{c + 2 \cdot d + 3 \cdot a} + \frac{c}{d + 2 \cdot a + 3 \cdot b} + \frac{d}{a + 2 \cdot b + 3 \cdot c} \ge \frac{2}{3}$

Oct 25, 2017

To prove any sort of equation or theorem, you plug in numbers and see if it's correct.

So the question is asking you to plug in random positive real numbers for a,b,c,d and see if the left expression is less than or equal to $\frac{2}{3}$.

Pick any random positive real numbers for a,b,c,d. 0 is a real number but it is neither positive or negative.
$a = 1 , b = 1 , c = 1 , d = 1$

$\frac{a}{b + 2 \cdot c + 3 \cdot d} + \frac{b}{c + 2 \cdot d + 3 \cdot a} + \frac{c}{d + 2 \cdot a + 3 \cdot b} + \frac{d}{a + 2 \cdot b + 3 \cdot c} \ge \frac{2}{3}$

Plug in numbers and simplify to see if it's greater or equal to the right expression.
$\frac{1}{1 + 2 \cdot 1 + 3 \cdot 1} + \frac{1}{1 + 2 \cdot 1 + 3 \cdot 1} + \frac{1}{1 + 2 \cdot 1 + 3 \cdot 1} + \frac{1}{1 + 2 \cdot 1 + 3 \cdot 1} \ge \frac{2}{3}$
$\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \ge \frac{2}{3}$
$\frac{2}{3} \ge \frac{2}{3}$
So with $a = 1 , b = 1 , c = 1 , d = 1$ it passes the inequality. This means that the domain for $a , b , c , d$ is from $1$ to $\infty$.