A math proof problem in the middle and high school bridging textbook

In fact, if a, b, c, and d are positive real numbers, the conclusion is true.

Because it can be proved by the mean inequality, when a, b, c, d>0, there must be a 4+b 4+c 4+d 4>=4abcd

And when a=b=c=d, the equal sign is taken.

Mean inequality: a1+a2+.an>=n*[n times the root number (a1*a2*...)an)]

If and only if a1=a2=....=an.

From the mean inequality, we can see that a 4+b 4+c 4+d 4>=4*fourth-order root number (a 4*b 4*c 4*d 4)=4abcd

So a 4 + b 4 + c 4 + d 4> = 4abcd is constant for a, b, c, d>0.

From the mean inequality, the equality sign can be established only when a=b=c=d, and the equal sign can be established.

In junior high school, you should have learned the binary mean inequality a+b>=2 ab

I can't help it if I haven't studied it, but high school requires mastery of n-yuan mean inequalities. That's how this problem is solved. You didn't go to junior high school?

Then your junior high school is too. I had already learned about n-yuan mean inequalities in junior high school

It is called"Empty gloves white wolf"Method:

a^4+b^4+c^4+d^4-4abcd(a^4-2a^2b^2+b^4)+(c^4+d^4-2c^2d^2)+2(a^2b^2+c^2d^2-2abcd)

A 2-b 2) 2+(c 2-d 2) 2+2(ab-cd) 2 that can. a^2=b^2

c^2=d^2

ab=cd, that is.

a=bc=da=c, i.e.

a=b=c=d

a b-b a-(a 2 + b 2) ab divides it into (a 2-b 2-a 2-b 2) ab and then simplifies it to -2b a

Because (a+b) 2=a 2+2ab+b 2, ab=[(a+b) 2-a 2-b 2] 2 is substituted into 3a 2+ab-2b 2=0 to obtain: 3a 2+[(a+b) 2-a 2-b 2] 2-2b 2=0

So [6a 2+(a+b) 2-a 2-b 2-4b 2] 2=0 so 6a 2+(a+b) 2-a 2-b 2-4b 2=0 so 5a 2-5b 2+(a+b) 2=0 so 5(a 2-b 2)+(a+b) 2=0 so 5(a+b)(a-b)+(a+b) 2=0 Situation: 1When a+b is not equal to 0, the formula divides a+b at the same time to obtain 5(a-b)+(a+b)=0

Deparentheses: 6a-4b=0

So b=3a2

So b a=3 2

So -2b a = -3

2.When a+b=0, there is a=-b

So -2b a=2

To sum up: a b-b a-(a 2 + b 2) ab has a value of -3 or 2

Solution: Decompose, got.

3a-2b）（a+b）=0

then a=-b or a=2 3b

When a=-b, substitute the original formula to obtain.

b b + b b - (b 2 + b 2) (-b) b = -1 + 1 + 2 = 2 when a = 2 3b, substitution is obtained.

2/3b/b-b/2/3b-(4/9b^2+b^2/)/2/3b×b=2/3-3/2-13/6=-3

In summary, the original value is 2 or -3

I don't understand, please ask, I wish you a happy o(o

3a²+ab-2b²=0

a+b）（3a-2b）=0

So. a+b=0 or 3a-2b=0

a=-b or a=2b3

When a=-b, the original formula =-1+1-(b +b) (-b )=2 When a=2b 3, the original formula = 2 3-3 2-(4b 9+b) (2b 3)=-3

Answer: A known question can be reduced to (3a-2b)(a+b)=0 because neither a nor b = 0

Therefore, 3a-2b=0 or a+b=0 then b a=3 2 or b a=-1 is the formula used: a b-b a-(a 2+b 2) ab homodifferentiation simple = -2b a, substituting the value of b a obtained above to obtain -3 or 2

The result is -3 or 2.

Solution: a, b are the two real roots of the equation 2x -4mx+2m +3m-2=0 δ=(-4m) -4 2 (2m +3m-2) 0 to get m2 3

From Veda's theorem:

a+b=2m，ab=(2m²+3m-2)/2∴a²+b²=2(m-3/4)²+7/8=2(3/4-m)²+7/8∵m≤2/3

3/4-m≥3/4-2/3＞0

When m=2 3, a +b obtains the minimum value as: 2 (3 4-2 3) +7 8=8 9

Because a and b are real roots, (-4m) -4*2*(2m +3m-2)>=0, the solution is m<=2 3

On the other hand, a + b = (a + b) -2ab, where a + b = 4m 2; ab = (2m + 3m-2) 2 solution a +b = 2 * (m-3 4) +7 8

According to the previous conditional constraint of m, when m=2 3, there is a minimum value of 8 9

It's all factoriable.

x²+2x+1-a²≤0

x+1-a)(x+1+a)≤0

Then you can discuss the positive and negative of a (that is, compare the size of 1-a and 1+a) a< 0 a-10 -a-11, 1 because it is the initial elevation, this kind of problem can be simply factored, pay attention to the observation and try to decompose it by multiplying the cross.

= (x + 1+a)(x + 1-a) ≤0-1-a ≤ x ≤ a-1

2.That is, (x-a)(x-1) <0, and the solution set is in the form x11,1 by comparing the sizes of a and 1

1x²+2x+1-a²≤0

x+1-a)(x+1+a)≤0

When a-1>-a-1 is a>0, a-1 x a-1

When a-1>-a-1 is a>0, a-1 x a-1

When a-1=-a-1 i.e. a=0, x=-a-1=a-1

2x²-（1+a）x+a＜0

x-1)(x-a)<0

When a=1, x has no solution.

When a<1, a1, 1

Criss-cross.

x (1-a)

x (1+a)

Solve x a-1, -a-1

x -1x -a

Solution x 1, a

(x+1) is less than or equal to the square of a.

x-1) (x-a) is less than 0 The following should be the main recipe Learn to multiply with crosses.

Solution: According to Veda's theorem, there is:

x1+x2=-p，x1•x2=q，(x1+1)(x2+1)=p,x1+1+x2+1=-q

According to x1+x2=-p, x1+1+x2+1=-q, there is: p-q=2 according to x1+x2=-p, x1 x2=q, (x1+1)(x2+1)=p has: 2p-q=1

Therefore: p=-1, q=-3

Therefore, b is selected according to m + m-4 = 0, 1 n + 1 n-4 = 0, and m ≠ 1 n so that m and 1 n can be regarded as two real numbers of x + x-4 = 0: m + 1 n = -1 so b is selected