Prove this!

One simple proof is when:

a=1 and b=1 !!.

Nice one guest.

good gest. vry nic.

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)

2-b³>8-b³+6b²-12b ........add (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+3b^2*c₂-3b*c₂² ---------->

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

0<=c₂³-c₁+3b²*c₂-3b*c_2²=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 c₁ is 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))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.