If r and s are natural numbers, then 1/2 (r+s)≥√rs. How can I proof it using direct proof? Thank you.
still on this?
square both sides and see what happens
oh, thanks oobleck
To prove the inequality 1/2 (r+s) ≥ √rs using direct proof, you need to follow these steps:
Step 1: Start with the statement to be proven.
1/2 (r+s) ≥ √rs
Step 2: Square both sides of the inequality to eliminate the square root.
(1/2 (r+s))^2 ≥ (√rs)^2
Step 3: Simplify both sides.
1/4 (r^2 + 2rs + s^2) ≥ rs
Step 4: Expand both sides.
(r^2 + 2rs + s^2)/4 ≥ rs
Step 5: Multiply both sides by 4 to eliminate the fraction.
r^2 + 2rs + s^2 ≥ 4rs
Step 6: Rearrange the terms.
r^2 - 2rs + s^2 ≥ 0
Step 7: Factor the left-hand side.
(r - s)^2 ≥ 0
Step 8: Analyze the inequality.
Since a square is always non-negative or equal to zero, (r - s)^2 ≥ 0 holds true for all values of r and s.
Step 9: Conclusion.
Therefore, 1/2 (r+s) ≥ √rs is true for any natural numbers r and s.
By following these steps, you have proven the inequality using direct proof.