- #1
maxkor
- 84
- 0
Let x,y>0 and x+y=2. Prove $\sqrt{x+\sqrt[3]{y^2+7}}+\sqrt{y+\sqrt[3]{x^2+7}}\geq2\sqrt3$
To prove this inequality, we can use the fact that if a + b = c, then a < c and b < c. So, in this case, if x + y = 2, then x < 2 and y < 2. Since x and y are both less than 2, x must be less than y.
Yes, you can use algebra to prove this inequality. One way to do so is by subtracting y from both sides of the equation x + y = 2, which gives us x = 2 - y. Then, we can substitute this value of x into the inequality x < y to get (2 - y) < y. By solving for y, we can see that this is true for all values of y that are less than 1.
Yes, there is a graphical way to prove this inequality. We can plot the points (x, y) where x + y = 2 on a coordinate plane and then draw a line connecting these points. This line will represent all possible values of x and y that satisfy the equation. We can then shade the region below this line to represent the inequality x < y. This visually shows that all points below the line satisfy the inequality.
Yes, you can prove this inequality by contradiction. Assume that x ≥ y, and then add x and y to get 2x ≥ 2. Since x and y are both positive, this means that x ≥ 1. But if x ≥ 1, then y ≤ 1, which contradicts the fact that x + y = 2. Therefore, our assumption must be false, and x < y.
Yes, this inequality is always true when x and y are positive and x + y = 2. This can be seen by looking at the graph of x + y = 2 and the line representing x < y. The line will always be below the graph, showing that x < y is always true for this equation.