If-a-and-b-positive-real-number-where-a-505-b-505-1-then-minimum-value-a-2020-b-2020-is-

Question Number 115033 by bobhans last updated on 23/Sep/20

I f a a n d b p o s i t i v e r e a l n u m b e r w h e r e

a^{505} + b^{505} = 1, t h e n m i n i m u m v a l u e

a^{2020} + b^{2020} i s__

Answered by 1549442205PVT last updated on 23/Sep/20

Put a^{505} = x, b^{505} = y \Rightarrow x + y = 1 . We need

find minimum value of P = x^{4} + y^{4}

Since x, y > 0, 1 = x + y = {(\sqrt{x} - \sqrt{y})}^{2} + 2 \sqrt{xy} ⩾ 2 \sqrt{xy}

\Rightarrow \sqrt{xy} ⩽ 1 / 2 \Rightarrow xy ⩽ 1 / 4 (1) . Hence,

x^{4} + y^{4} = {(x + y)}^{4} - 4 xy (x^{2} + y^{4}) - {6 x}^{2} y^{2}

= 1 - 4 xy [{(x + y)}^{2} - 2 xy] - {6 x}^{2} y^{2}

= 1 - 4 xy (1 - 2 xy) - 6 {(xy)}^{2} = 1 - 4 xy + 2 {(xy)}^{2}

= 2 {(xy - \frac{1}{4})}^{2} - 3 xy + \frac{7}{8} ⩾ 0 - 3 . \frac{1}{4} + \frac{7}{8}

= \frac{1}{8} . The equality ocurrs if and only

if x = y = 1 / 2 \Leftrightarrow a = b =^{505} \sqrt{\frac{1}{2}}

Thus {(a^{2020} + b^{2020})}_{\min} = \frac{1}{8}

when a = b = \frac{1}{^{505} \sqrt{2}}

Answered by bemath last updated on 23/Sep/20

Facebook Tweet Pin