Menu Close

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
     Find the minimum value of k       such that for arbitrary a,b >0       we have  (a)^(1/(3 ))  + (b)^(1/(3 ))  ≤ k ((a+b))^(1/(3 ))
$$\:\:\:\:\:{Find}\:{the}\:{minimum}\:{value}\:{of}\:{k} \\ $$$$\:\:\:\:\:{such}\:{that}\:{for}\:{arbitrary}\:{a},{b}\:>\mathrm{0} \\ $$$$\:\:\:\:\:{we}\:{have}\:\:\sqrt[{\mathrm{3}\:}]{{a}}\:+\:\sqrt[{\mathrm{3}\:}]{{b}}\:\leqslant\:{k}\:\sqrt[{\mathrm{3}\:}]{{a}+{b}}\: \\ $$
Answered by EDWIN88 last updated on 18/May/21
 consider the function f(x) = (x)^(1/(3 ))    we have f ′(x)= (1/3) x^(−(2/3))  & f ′′(x)=−(2/9)x^(−(5/3))    for any x ∈ (0,∞) thus f(x) concave on the  interval (0,∞).  By Jensen′s inequality    we deduce             (1/2)f(a) + (1/2)f(b) ≤ f (((a+b)/2))            (((a)^(1/(3 ))  + (b)^(1/(3 )) )/2) ≤ (((a+b)/2))^(1/(3 ))            (a)^(1/(3 ))  + (b)^(1/(3 ))  ≤ ((2 ((a+b))^(1/(3 )) )/( (2)^(1/(3 )) )) = ((2 (4)^(1/3) )/2)(a+b)         minimum value of k is (4)^(1/(3 ))  ⋇
$$\:\mathrm{consider}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt[{\mathrm{3}\:}]{\mathrm{x}} \\ $$$$\:\mathrm{we}\:\mathrm{have}\:\mathrm{f}\:'\left(\mathrm{x}\right)=\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{x}^{−\frac{\mathrm{2}}{\mathrm{3}}} \:\&\:\mathrm{f}\:''\left(\mathrm{x}\right)=−\frac{\mathrm{2}}{\mathrm{9}}\mathrm{x}^{−\frac{\mathrm{5}}{\mathrm{3}}} \\ $$$$\:\mathrm{for}\:\mathrm{any}\:\mathrm{x}\:\in\:\left(\mathrm{0},\infty\right)\:\mathrm{thus}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{concave}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{interval}\:\left(\mathrm{0},\infty\right).\:\:\mathrm{By}\:\mathrm{Jensen}'\mathrm{s}\:\mathrm{inequality}\: \\ $$$$\:\mathrm{we}\:\mathrm{deduce}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{f}\left(\mathrm{a}\right)\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{f}\left(\mathrm{b}\right)\:\leqslant\:\mathrm{f}\:\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{a}}\:+\:\sqrt[{\mathrm{3}\:}]{\mathrm{b}}}{\mathrm{2}}\:\leqslant\:\sqrt[{\mathrm{3}\:}]{\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\sqrt[{\mathrm{3}\:}]{\mathrm{a}}\:+\:\sqrt[{\mathrm{3}\:}]{\mathrm{b}}\:\leqslant\:\frac{\mathrm{2}\:\sqrt[{\mathrm{3}\:}]{\mathrm{a}+\mathrm{b}}}{\:\sqrt[{\mathrm{3}\:}]{\mathrm{2}}}\:=\:\frac{\mathrm{2}\:\sqrt[{\mathrm{3}}]{\mathrm{4}}}{\mathrm{2}}\left(\mathrm{a}+\mathrm{b}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:{k}\:\mathrm{is}\:\sqrt[{\mathrm{3}\:}]{\mathrm{4}}\:\divideontimes \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *