Question

⁴√ab(√a+√b)≤a+b доказать при a≥0 b≥0

Answer 1

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Answer 2

To prove the inequality ⁴√(ab)(√a + √b) ≤ a + b for a ≥ 0 and b ≥ 0, we'll start by squaring both sides of the inequality to eliminate the fourth root:

Step 1: Square both sides (⁴√(ab)(√a + √b))^2 ≤ (a + b)^2

Step 2: Simplify the left side (ab)(√a + √b)^4 ≤ (a + b)^2

Step 3: Expand the left side (ab)(a^2 + 2a√a√b + b^2) ≤ (a + b)^2

Step 4: Expand the left side again a^3b + 2a^2b√ab + ab^3 ≤ a^2 + 2ab + b^2

Step 5: Move all terms to the left side a^3b + 2a^2b√ab + ab^3 - a^2 - 2ab - b^2 ≤ 0

Now, we will prove the inequality step by step:

Step 1: Show that (a - b)^2 ≥ 0 This is always true since the square of any real number is non-negative. Therefore, (a - b)^2 ≥ 0.

Step 2: Expand (a - b)^2 (a - b)^2 = a^2 - 2ab + b^2

Step 3: Rearrange the terms a^2 - 2ab + b^2 = a^2 + b^2 - 2ab

Step 4: Show that a^2 + b^2 ≥ 2ab Since a ≥ 0 and b ≥ 0, both a^2 and b^2 are non-negative. So, a^2 + b^2 ≥ 0. From Step 1, we know that (a - b)^2 ≥ 0, which means a^2 + b^2 - 2ab ≥ 0. Therefore, a^2 + b^2 ≥ 2ab.

Step 5: Substitute a^2 + b^2 ≥ 2ab into the original inequality a^3b + 2a^2b√ab + ab^3 - a^2 - 2ab - b^2 ≤ a^2 + b^2 - 2ab

Step 6: Simplify a^3b + 2a^2b√ab + ab^3 - a^2 - 2ab - b^2 + 2ab - a^2 - b^2 ≤ 0

Step 7: Cancel out some terms a^3b + ab^3 + 2a^2b√ab - b^2 ≤ 0

Step 8: Factor out ab ab(a^2 + b^2 + 2√ab) - b^2 ≤ 0

Step 9: Factor a^2 + b^2 + 2√ab as (a + b)^2 ab(a + b)^2 - b^2 ≤ 0

Step 10: Factor out b^2 from the left side b^2[ab(a + b)^2/b^2 - 1] ≤ 0

Step 11: Divide through by ab(a + b)^2/b^2 b^2/b^2 - 1 ≤ 0

Step 12: Simplify 1 - 1 ≤ 0

Since 0 ≤ 0 is true, this proves that the original inequality is also true:

⁴√(ab)(√a + √b) ≤ a + b for a ≥ 0 and b ≥ 0.