Let-p-q-r-are-positive-real-numbers-0-lt-r-lt-min-p-q-Prove-that-p-r-q-r-min-pq-r-2-p-q-2r-

Question Number 71206 by naka3546 last updated on 13/Oct/19

L e t p, q, r a r e p o s i t i v e r e a l n u m b e r s .

0 < r < m i n {p, q} .

P r o v e t h a t

\sqrt{p - r} + \sqrt{q - r} ⩽ m i n {\sqrt{\frac{p q}{r}}, \sqrt{2 (p + q - 2 r)}}

Answered by mind is power last updated on 13/Oct/19

\sqrt{p - r} + \sqrt{q - r} =⩽ \sqrt{2 (p + q - 2 r)}

\sqrt{x} + \sqrt{y} ⩽ \sqrt{2 x + 2 y}

cause x + y - 2 \sqrt{xy} = {(\sqrt{x} - \sqrt{y})}^{2} ⩾ 0

\Rightarrow x + y + 2 \sqrt{xy} ⩽ 2 x + 2 y

\Rightarrow \sqrt{x} + \sqrt{y} ⩽ \sqrt{2 (x + y)}

\Rightarrow x = p - r y = q - r

\sqrt{p - r} + \sqrt{q - r} ⩽ \sqrt{2 (p + q - 2 r)} \dots . 1

\sqrt{p - r} + \sqrt{q - r} ⩽ \sqrt{\frac{pq}{r}}

\sqrt{p - r} + \sqrt{q - r} = \sqrt{r} \sqrt{\frac{p}{r} - 1} + \sqrt{r} \sqrt{\frac{q}{r} - 1}

\sqrt{x - 1} + \sqrt{y - 1} ⩽ \sqrt{xy}

x + y - 2 + 2 \sqrt{(x - 1) (y - 1)} ⩽ xy

2 \sqrt{(x - 1) (y - 1)} ⩽ xy - x - y + 2 = (x - 1) (y - 1) + 1

\Rightarrow {(\sqrt{(x - 1) (y - 1)} - 1)}^{2} ⩾ 0 True

x = \frac{p}{r}, y = \frac{q}{r} \Rightarrow

\sqrt{\frac{p}{r} - 1} + \sqrt{\frac{q}{r} - 1} ⩽ \sqrt{\frac{pq}{r^{2}}}

\Rightarrow \sqrt{r} (\sqrt{\frac{p}{r} - 1} + \sqrt{\frac{q}{r} - 1}) ⩽ \sqrt{\frac{pq}{r}} \dots 2

2 & 1 \Rightarrow

\sqrt{p - r} + \sqrt{q - r} ⩽ \min {\sqrt{\frac{pq}{r}}, \sqrt{2 (p + q - 2 r)}}