Getting close to the minimum

Algebra Level 3

a , b a,b are reals,then what is the minimum value of a 2 + b 2 + a b + a + b + 1 a^2+b^2+ab+a+b+1 ?

Doesn't exist 1 3 \frac13 2 3 \frac23 1 0

X X by X X , 17, Taiwan

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2 solutions

X X
May 18, 2018

Let a = c + d , b = c − d a=c+d,b=c-d ,then a 2 + b 2 + a b + a + b + 1 = 3 c 2 + d 2 + 2 c + 1 = 3 ( c + 1 3 ) 2 + d 2 + 2 3 a^2+b^2+ab+a+b+1=3c^2+d^2+2c+1=3(c+\frac13)^2+d^2+\frac23 ,so the minimum value is 2 3 \frac23 .It happens when c + 1 3 = 0 , d = 0 c+\frac13=0,d=0

1 Helpful 1 Interesting 2 Brilliant 0 Confused
Aaghaz Mahajan
May 16, 2018

Simply use Partial Derivatives and see that minimum occurs at a=b=(-1/3).............

1 Helpful 0 Interesting 0 Brilliant 0 Confused

You still have to show that this critical point is a global minimum.

Pi Han Goh - 3 years ago

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