Let-a-b-c-R-and-a-b-b-c-1-where-0-lt-b-1-Prove-that-a-b-b-c-a-b-b-c-2-

Question Number 78351 by loveineq. last updated on 16/Jan/20

Let a, b, c \in R^{+} and (a + b) (b + c) = 1, where 0 < b ⩽ 1 .

Prove that ∣ a - b ∣∣ b - c ∣ ⩾ \frac{∣ \sqrt{a} - \sqrt{b} ∣∣ \sqrt{b} - \sqrt{c} ∣}{2} .

Commented by loveineq. last updated on 19/Jan/20

More stronger is ∣ a - b ∣∣ b - c ∣ ⩾ ∣ \sqrt{a} - \sqrt{b} ∣∣ \sqrt{b} - \sqrt{c} ∣ .

Answered by mind is power last updated on 16/Jan/20

x - y = (\sqrt{x} - \sqrt{y}) (\sqrt{x} + \sqrt{y})

\Rightarrow∣ a - b ∣∣ b - c ∣=∣ \sqrt{a} - \sqrt{b} ∣∣ \sqrt{b} - \sqrt{c} ∣ (\sqrt{a} + \sqrt{b}) (\sqrt{b} + \sqrt{c})

\sqrt{x} + \sqrt{y} ⩾ \frac{\sqrt{x + y}}{\sqrt{2}} proof \Leftrightarrow 2 (x + y + 2 \sqrt{xy}) ⩾ x + y

\Leftrightarrow x + y + 4 \sqrt{xy} ⩾ 0 True

(\sqrt{a} + \sqrt{b}) (\sqrt{b} + \sqrt{c}) ⩾ \frac{\sqrt{a + b} . \sqrt{b + c}}{2} = \frac{\sqrt{(a + b) (b + c)}}{2} = \frac{1}{2}

\Rightarrow∣ \sqrt{a} - \sqrt{b} ∣∣ \sqrt{b} - \sqrt{c} ∣ (\sqrt{a} + \sqrt{b}) (\sqrt{b} + \sqrt{c}) ⩾ \frac{∣ \sqrt{a} - \sqrt{b ∣} ∣ \sqrt{b} - \sqrt{c} ∣}{2}

Answered by loveineq. last updated on 16/Jan/20

Thanks sir

Commented by mind is power last updated on 19/Jan/20

y^{'} re Welcom