Find-the-minimum-value-of-k-such-that-for-arbitrary-a-b-gt-0-we-have-a-1-3-b-1-3-k-a-b-1-3-

Question Number 141405 by iloveisrael last updated on 18/May/21

F i n d t h e m i n i m u m v a l u e o f k

s u c h t h a t f o r a r b i t r a r y a, b > 0

w e h a v e \sqrt[3]{a} + \sqrt[3]{b} ⩽ k \sqrt[3]{a + b}

Answered by EDWIN88 last updated on 18/May/21

consider the function f (x) = \sqrt[3]{x}

we have f^{'} (x) = \frac{1}{3} x^{- \frac{2}{3}} & f^{″} (x) = - \frac{2}{9} x^{- \frac{5}{3}}

for any x \in (0, \infty) thus f (x) concave on the

interval (0, \infty) . By {Jensen}^{'} s inequality

we deduce

\frac{1}{2} f (a) + \frac{1}{2} f (b) ⩽ f (\frac{a + b}{2})

\frac{\sqrt[3]{a} + \sqrt[3]{b}}{2} ⩽ \sqrt[3]{\frac{a + b}{2}}

\sqrt[3]{a} + \sqrt[3]{b} ⩽ \frac{2 \sqrt[3]{a + b}}{\sqrt[3]{2}} = \frac{2 \sqrt[3]{4}}{2} (a + b)

minimum value of k is \sqrt[3]{4} ⋇

Facebook Tweet Pin