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Question Number 173981 by AgniMath last updated on 22/Jul/22

Answered by Frix last updated on 22/Jul/22

a≥0∧b≥0 ⇒ we can square (((a+b)^2 )/4)≥ab (a+b)^2 ≥4ab a^2 +2ab+b^2 ≥4ab a^2 −2ab+b^2 ≥0 (a−b)^2 ≥0 true

a⩾0∧b⩾0⇒wecansquare(a+b)24⩾ab(a+b)2⩾4aba2+2ab+b2⩾4aba2−2ab+b2⩾0(a−b)2⩾0true

Answered by puissant last updated on 22/Jul/22

((√a)−(√b))^2 ≥0 ⇔ a+b−2(√a).(√b) ≥0 ⇔ a+b ≥ 2(√(ab)) ⇔ ((a+b)/2) ≥ (√(ab))

(a−b)2⩾0⇔a+b−2a.b⩾0⇔a+b⩾2ab⇔a+b2⩾ab

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