Verification The value of the polynomial A 2 b 2 2a 10 is always non negative, regardless of the val

Proof: A2+B 2-2A+10

a^2-2a+1)+(b^2+9)

a-1)^2+(b^2+9)

Because (a-1) 2 0 and b 2+9>0

So (a-1) 2+(b 2+9) >0

That is, the polynomial A 2 + B 2-2A + 10>0 is constant.

So no matter what the value is, the value of the polynomial a 2 + b 2-2a + 10 is always non-negative.

Proof: a 2 + b 2-2a + 10 = (a - 1) + b + 9

Because (a-1) 0, b 0, so (a-1) +b +9 0, so a 2 + b 2-2a + 10 0

So no matter what the value is, the value of the polynomial a 2 + b 2 - 2a + 10 is always greater than 0 and is not negative.

No way? First split 10 into 9 and 1 Then the formula can know that this polynomial must be a non-negative number Formula: (a-1) 2+b 2+9 This polynomial is evergreen to 0!

Let a and b be non-negative real numbers, then the sufficient condition for a+b<5 4 is () 1 ab<1 16 2 a 2+b 2<1

Proof: Method: With a as the main yuantong sliding call, the secondary correspondence bureau Kai number is constructed, and the discriminant method is used.

To prove f(a)=a 2 (b-1)a b 2-b 1>0, it is only necessary to prove that it is a positive distinction <0. Its discriminant formula =(b-1) 2-4(b 2-b 1)=b 2-2b 1-4b 2 4b-4 =-3b 2 2b-3=-3(b 2-2b 3)-3=-3[(b-1 3) 2-1 9]-3 =-3(b-1 3) 2-8 3<0 holds for any b. The image of f(a) is a parabola with b as a parametric and the opening is facing upwards, and for any parameter b, its discriminant formula is 0, i.e., the inequality a 2 b 2 ab 1 > a b is proved.

Original formula = a 2b 2+a 2+b 2-4ab[(ab) 2-2ab+1]+(a 2+b 2-2ab)+1(ab-1) 2+(a-b) 2+1

You can hold the imperial branch to see that the original form is always Duan Min non-negative.

2a+b=1

b = 1-2a, w=2a 2 (a 2+b)+1 b 22a 2 (a 2-2a+1)+1 (1-2a) 22a 2 (a-1) 2+1 (2a-1) 2,0 a<1 2,w'Excite car = 4a (a-1) 2-4a 2 (a-1) 3-4 (2a-1) 3

4a (a-1) 3-4 (2a-1) 3> Lub 0, so w is the increasing function, a=0 when b=1, w=1, is the minimum value sought.

It is obtained from a, b, c as non-negative numbers and the square mean is greater than the high bucket and equal to the arithmetic mean.

a*a+b*b) 2) (a+b) 2((b*b+c*c) 2) b+c) 2( a*a+c*c ) 2) a+c) 2) a+c) 2 The above three formulas can be proved by adding the above three formulas and dividing Qi this grinding root 2 to the right.

Condition 2 can be seen as a point in a circle, and finding the maximum value of a+b = root number 2>5 4 is insufficient.

or a+b=root(a2+b2+2ab)<=root[2(a2+b2)]=root2

Condition 1 derives a=1 16b so that its reciprocal is equal to minus 1 to give a=b=1 4, a+b<5 4 , sufficient.

If a +b <1, then ab<1 2,a+b) <1+1=2

a+b<√2

But 2>5 4, so a + b <1 is not a sufficient condition.

If one is excluded, the other is enough, so ab<1 16 is a sufficient condition.

|a-b| +ab = 1

A and b are non-negative integers, then.

1. |a-b|= 0, ab = 1, a = b = 12 |a-b| = 1, ab =0

a = 0, b =1 (b = -1 rounded).

or b = 0, a =1 (a=-1 rounded) so there are three sets of solutions: a = b = 1; a = 0, b = 1; a =1, b = 0

Solution: Because 2001=(a+1)(b+1)

And 2001 = 1 * 2001 = 3 * 667

So the values of a and b can be .,or 2,666

So a+b=0+2000=2000, or a+b=2+666=668