Senior 1 math competition questions, advanced masters, mathematics. High. Olympiad questions. A grea

a^4+b^4

a^2+b^2)^2-2a^2*b^2

a+b)^2-2ab]^2-2a^2*b^2(1-2ab)^2-2a^2*b^2

1+2a^2b^2-4ab

2(ab-1)^2-1

Because a+b=1 a>0, b>0

So ab<=[(a+b) 2] 2=1 4, so when the minimum value of a 4+b 4 is ab=1 4, a 4+b 4=2(ab-1) 2-1=1 8

So 2x 2-x+a 4+b 4>=2x 2-x+1 8 because 2x 2-x+1 8=2(x-1 4) 2>=0 holds for any real number x.

So 2x 2-x+a 4+b 4>=2x 2-x+1 8>=0 holds for any real number x.

The square of all real numbers (a+b) = 4ab 1>=16a squared * b squared 0<=a square b square< = 1 16 with x equation solution (2x square-x+1) 6>=a square b square (2x square-x+1) 6>= 1 16 16x square-8x=5>=0 deter<0 parabola above the x-axis is the whole real number.

Solve the two roots, and x should be larger than the big root or smaller than the small root.

1) Solution: The quadratic function satisfies f(x+1)-f(x)=2x imaginary sum a(x+1) +b(x+1)+c-(ax +bx+c)=2ax+a+b=2x

That is, 2a=2, a+b=0

a=1，b=-1

and f(0)=1

That is, it is worth substituting c = 1 into a, b, c.

f(x)=x²-x+1

2) Solution: Inequality f(x)>2x+m simplification:

x²-3x+1>m

For any x belongs to [-1; 1], the inequality f(x)>2x+m constant difference staring.

f(x) is a subtractive function at x [-1,1].

So if the inequality is constant.

i.e. f(1)-3x1>m

Solution: m<-1

Mathematical induction.

or [sina + sin (a + b) + sin (a + 2b) + .]sin(a+nb)]sin(b/2)

sinasin(b/2)+sin(a+b)sin(b/2)+sin(a+2b)sinb/2)+.sin(a+nb)sin(b/2)

-1/2)[cos(a+b/2)-cos(a-b/2)+cos(a+3b/2)-cos(a+b/2)+.cos(a+(2n+1)b/2)-cos(a+(2n-1)b/2)

-1/2)[cos(a+(2n+1)b/2)-cos(a-b/2)]

sin(a+nb/2)sin(n+1)b/2

i.e. sina + sin (a + b) + sin (a + 2b) +sin(a+nb)=sin(a+ab/2)sin[(n+1)b/2]/sin(b/2)

Give a method, use mathematical induction.

Let x1=k(x2+x3+x4).

1 3 (x2 + x3 + x4) < = x1< = x2 + x3 + x4 then 1 3< = k< = 1

The original inequality is deformed as.

1+k)^2(x2+x3+x4)^2<=4k(x2+x3+x4)x2x3x4

1+k)^2/4k](x2+x3+x4)<=x2x3x4①[(1+k)^2/4k](x2+x3+x4)<=[(1+k)^2/4k](x2+x2+x2)=[(1+k)^2/4k]*3x2

x2x3x4>=2*2*x2=4x2

Proof of establishment only needs to be proved.

1+k)^2/4k]*3x2<=4x21/4(k+1/k+2)*3<=4

Because f(x)=x+1 x is a subtractive function on [1 3,1], 1 4(k+1 k+2)*3<=(1 4)*(1 3+3+2)*3=4

Therefore (x1+x2+x3+x4) 2<=4x1x2x3x4

y^2+ay+b =2x^2+2x+c

>c=y^2+ay+b -(2x^2+2x)

It is only necessary to prove that y 2+ay+b -(2x 2+2x) cannot take all integers at x,y z.

That is, it proves that y 2+ay -(2x 2+2x) cannot take all integers at x, y z.

That is to say, it is clear that (y+1)(y+(a-1))-2 x(x+1) cannot take all integers.

Obviously, if a is an odd number, (y+1)(y+(a-1))-2 x(x+1) is always even, which obviously satisfies the conclusion in the title.

If a is an even number, you may wish to set a=2k, which is y 2 +2ky -2 x(x+1)=(y+k) 2 -2x(x+1)-k 2

In this case, it is enough to explain that (y+k) 2 -2x(x+1) cannot take all integers.

Now the inverse conclusion is not valid, that is, for any integer t=(y+k) 2 -2x(x+1) has a solution, that is, for any integer t, there is always an integer x, so that t+2x(x+1) is a square number.

Now assuming t=4m+3, then for the indefinite equation 4m+3+2x(x+1)=r 2, it is clear that r is an odd number.

Assuming r=2s+1, we get 4m+2=4s(s+1)-2x(x+1).

The left side is divided by 4 and 2, and the right side is a multiple of 4, which is the misunderstanding of this indefinite equation.

Therefore, the original conclusion is valid.

It can be seen as two parabolas, and the opening of 2x 2 is larger, so as long as it is assumed that the lowest point of M2 is directly below the lowest point of M1, then the two will never intersect, that is, for any A and B, C can always be found so that "the lowest point of M2 is directly below the lowest point of M1", which can satisfy the condition.