Final answer:
To prove x/(y+z) + y/(x+z) + z/(x+y) ≥ 3/2 for positive x, y, and z, multiply by (x+y+z), apply Cauchy-Schwarz inequality, and simplify to obtain 1/(y+z) + 1/(x+z) + 1/(x+y) ≥ 3/(x+y+z). As x, y, and z are positive, multiply both sides by (x+y+z) to establish the desired inequality.
Step-by-step explanation:
We want to prove that for any positive numbers x, y, and z, the expression x/(y+z) + y/(x+z) + z/(x+y) is always greater than or equal to 3/2. We can start by assuming that x, y, and z are positive.
First, we multiply the entire expression by (x+y+z), resulting in (x^2 + y^2 + z^2)/(y+z) + (x^2 + y^2 + z^2)/(x+z) + (x^2 + y^2 + z^2)/(x+y) ≥ 3/2 * (x+y+z).
Next, we use the Cauchy-Schwarz inequality to simplify. This leads to ((x^2 + y^2 + z^2)(1/(y+z) + 1/(x+z) + 1/(x+y))) ≥ (x+y+z)^2 / (x+y+z).
We then simplify further and divide both sides by (x+y+z), which gives us 1/(y+z) + 1/(x+z) + 1/(x+y) ≥ 3/(x+y+z).
Since x, y, and z are positive, the denominators of each fraction are positive.
Therefore, we can multiply both sides of the inequality by (x+y+z) and simplify to obtain the desired inequality x/(y+z) + y/(x+z) + z/(x+y) ≥ 3/2.
For positive numbers X,Y,Z we have Z/Y= 35/6 and Z/X=35/8. Column A X Column B Y
*It is a mistake to assume that Z = 35 and X = 8 and Y = 6* These are all ratios, so the best we can assume is that Z = 35k, X= 8k, and Y= 6k. And so we notice that 8k is always going to be bigger for any positive number than 6k.