Proving the Inequality a(a+b)+b(b+1) ≥ -1/3

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[MrLidless]

Partial differentials gives 2a+b = 0, 2b+a+1 = 0. Simple to solve from there. Partially differentiating again shows that these are minimums.

[MizardXYT]

Expanding gives a² + ab + b² + b
There is a cross-term (ab), so substitute a=u + v, b=u - v giving 3u² + u + v² - v
Complete the squares in both u and v gives 3(u + 1/6)² + (v - 1/2)² - 1/3
We can already see that this is ≥ -1/3, but let's continue anyway.
Substituting back with u=(a + b)/2, v=(a - b)/2 gives 3(a/2 + b/2 + 1/6)² + (a/2 - b/2 - 1/2)² - 1/3
We can verify this by expanding into a² + ab + b² + b

[kpt]

nice solution both methods people generally underestimate the statement that no (real) square is negative. well done. keep doing

[MathTutor]

I think there are two different problems given. One in the title is different from the one in the cover page. In the solved version, the LHS is equal to (a+b/2)² + 3/4(b+2/3)² − 1/3, which leads to the answer. Note: This is actually similar to what he did. Motive: Whenever you see x and x² it is good to write it as a perfect square, if possible. Thank you.

[robertgerbicz]

Expression is (a+b/2)^2+1/3*(3/2*b+1)^2-1/3
from this the minimum is -1/3 and you can reach this only at a=1/3, b=-2/3.

[satyapalsingh]

Your method of solving the problem is praiseworthy, dear professor.

[frentz]

Completing the square gives X^2 + (3/4) * Y^2 - 1/3 for the left hand side, with X = a + b/2 and Y = b + 2/3

[robertodiasfb]

Such a nice and funny problem. Thanks!

[italixgaming]

My solution (of course way faster):
Let's define f(a, b)=a.(a+b)+b.(b+1). We're looking for the minimum of this function.
df=(2a+b).da+(2b+1+a).db
So we see that we have an extremum for 2a+b=0 and 2b+1+a=0. We can already see that this extremum is a minimum since, for example, f(0, b) can be as high as we want.
From the first equation we have b=-2a, we report into the second one and we have a=1/3 then b=-2/3.
Our value is: 1/3.(-1/3)-2/3.1/3=-1/3.

[jonathanward]

My motive for watching this site is to learn strategies and tactics in problem solving not only in math but problems in general. Thanks for the tactic multiplying by 4. That was good. By the way I solved this by the first method by watching previous videos.

[AnmolTheMathSailor]

Hey syber, again super coll problem. But thumbnail and tilte not equal 🙃

[cmagicalex]

You can try to use differentiation to find the extreme value

[thaity]

maybe we can start by saying a>b so a(a+b)+b(b+1)<3b²+b and there is no solution for 3x²+x+1/3=0 so its done

[-basicmaths]

Nice example!
This type of examples are based on algebraic principal if x and y are real numbers then x^2+y^2 is always greater than or equal to zero.
(a+b/2)^2+(√3/2b+1/√3)^2>=0

[jmart]

Great video. I solved it via 1st method.

[yoav]

Very nice video i liked your 2 methods well done😃

[explainingphysicsandmathematic]

Can you recommend a book that deeply explains the modern physics mathematics ?

[stewartcopeland]

@ 2:13 after 2b or not 2b, 4nite the Game ?

[epimaths]

Bất đẳng thức này nhìn là biết giải theo hằng đẳng thức.

[Timeflow_X]

cool video! just to clarify, are a and b just reals with no additional restrictions?