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# Prove this!

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Prove this :

If $$a^{3}+b^{3}=2$$ then $$a+b\leq 2$$

ant101  Apr 2, 2017
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#1
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One simple proof is when:

a=1 and b=1 !!.

Guest Apr 2, 2017
#2
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Nice one guest.

MysticalJaycat  Apr 2, 2017
#3
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good gest. vry nic.

mathtoo  Apr 2, 2017
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Lets assume a+b>2

a>2-b

Therefore,

a3>(2-b)3=8-b3+6b2-12b

We know a3=2-b3 (because a3+b3=2)

0>6+6b2-12b..........divide by 6----------->

0>1+b2-2b=(b-1)2

but (b-1)2 cant be smaller than 0!

we could also find the smallest number that is bigger than a+b if we called 2 'c1' and the limit for the largest number 'c2' to get a more general solution for every c1.

a3+b3=c1

a3=c1-b3

we want to find the smallest c2 that is bigger than or equals to a+b. (a+b<=c2)

a<=c2-b -------->

a3<=c23-b3+3b2*c2-3b*c22 ---------->

c1-b3<=c23-b3+3b2*c2-3b*c22 --------->

0<=c23-c1+3b2*c2-3b*c22=3c2*[(b-(c2/2))2+c22/3-c1/c2*3-c22/4]

so now there are 3 options-

1. c2 is negative, therefore we want [(b-(c2/2))2+c22/3-c1/c2*3-c22/4] to always be negative (because negative*negative=positive) but we cant because b can be as big as we want. meaning we cant do that.

2. c2=0, meaning c1=0. but cis a constant that cant be determined by c2.

3. c2 is a positive number, meaning we want [(b-(c2/2))2+c22/3-c1/c2*3-c22/4] to be positive always (because positive*positive=positive)

we know the minimum value for (b-(c2/2)) is 0. so we need c22/3-c1/c2*3-c22/4 to be bigger than 0.

c22/3-c1/c2*3-c22/4=c22/12-c1/c2*3>=0 multiply by 3*c2-------------->

c23*1/4-c1>=0

therefore, c2>=(c1*4)1/3

we can also use calculus for this question

a=x, b=(c1-x3)1/3

f(x)=x+(c1-x3)1/3

f'(x)=1-3*x2/3*(c1-x3)1/3-1=1-x2(c1-x3)-2/3=0 ----------------->

x2(c1-x3)-2/3=1 multiply by (c1-x3)2/3 -------------------->

x2=(c1-x3)2/3---------------------->

x6=(c1-x3)2=c12+x6-2*c1*x3------------------>

2*c1*x3=c12 divide by (c1*2) ------------->

x=(c1/2)1/3

f((c1/2)1/3)=(c1/2)1/3+(c1-c1/2)1/3=2*(c1/2)1/3. this is a more accurate answer for the maximum value of a+b.

Ehrlich  Apr 2, 2017
edited by Ehrlich  Apr 3, 2017

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