MHB Determine the largest number k

[anemone]

Determine the largest number $k$ such that system $a^2+b^2=1$, $|a^3-b^3|+|a-b|=k^3$ has a solution.

 

[Albert1]

Determine the largest number $k$ such that system $a^2+b^2=1$, $|a^3-b^3|+|a-b|=k^3---(1)$ has a solution.

Spoiler

from (1) we know that :$k\geq 0$
if $a=b$ then $k=0$
if $a>b$, we have :
$(a-b)(ab+2)=k^3=f(a,b)---(*)$
using lagrange-method:
we want to find:
$(a-b)(ab+2)+L(a^2+b^2-1)=max(f(a,b))$
we get $L=\dfrac {b-a}{2}=\dfrac {4ab+3}{2(b-a)}$
$\therefore ab=\dfrac {-1}{3}---(2)$
$a^2+b^2=1---(3)$
from (2)(3)$a-b=\sqrt{\dfrac {5}{3}}---(4)$
put (2)(4) to (*) and we get:
$k^3=\dfrac{5}{3}\times\sqrt{\dfrac{5}{3}}$
$k=\sqrt{\dfrac{5}{3}}$
if $b>a$
because of symmetry it will be the same

 

[anemone]

Spoiler

from (1) we know that :$k\geq 0$
if $a=b$ then $k=0$
if $a>b$, we have :
$(a-b)(ab+2)=k^3=f(a,b)---(*)$
using lagrange-method:
we want to find:
$(a-b)(ab+2)+L(a^2+b^2-1)=max(f(a,b))$
we get $L=\dfrac {b-a}{2}=\dfrac {4ab+3}{2(b-a)}$
$\therefore ab=\dfrac {-1}{3}---(2)$
$a^2+b^2=1---(3)$
from (2)(3)$a-b=\sqrt{\dfrac {5}{3}}---(4)$
put (2)(4) to (*) and we get:
$k^3=\dfrac{5}{3}\times\sqrt{\dfrac{5}{3}}$
$k=\sqrt{\dfrac{5}{3}}$
if $b>a$
because of symmetry it will be the same

Thanks for participating and well done, Albert! Your answer is correct! Something tells me you really like to use the AM-GM to solve for problem such as this one and for your information, the solution that I have used the AM-GM to solve it as well...do you think you want to give AM-GM a try? Hehehe...

 

[Albert1]

Thanks for participating and well done, Albert! Your answer is correct! Something tells me you really like to use the AM-GM to solve for problem such as this one and for your information, the solution that I have used the AM-GM to solve it as well...do you think you want to give AM-GM a try? Hehehe...

Spoiler

use AM-GM
$(a-b)^2=1-2ab$
$k^3=(a-b)(ab+2)=\sqrt {1-2ab}\times (ab+2)---(*)\leq\dfrac{(1-2ab)+(ab+2)^2}{2}$
$\therefore 1-2ab=ab+2,$ or $ab=\dfrac {-1}{3}$
put $ab=\dfrac {-1}{3} $to (*)we get:
$k=\sqrt {\dfrac {5}{3}}$

 

[anemone]

Spoiler

use AM-GM
$(a-b)^2=1-2ab$
$k^3=(a-b)(ab+2)=\sqrt {1-2ab}\times (ab+2)---(*)\leq\dfrac{(1-2ab)+(ab+2)^2}{2}$
$\therefore 1-2ab=ab+2,$ or $ab=\dfrac {-1}{3}$
put $ab=\dfrac {-1}{3} $to (*)we get:
$k=\sqrt {\dfrac {5}{3}}$

That's is it!(Yes) Thanks Albert for your second solution! :)