When p 1, p is a real number, a 0, b 0, a b p a p b p 20 is verified

Upstairs is not right, you must p as an integer to be true.

Constructor f(x) = (1+x) p - x p - 1 f'(x)= p(1+x)^(p-1) -p*x^(p-1) >p*（x)^(p-1) -p*x^(p-1) = 0

So in x>= 0, f(x) is strictly incremented. And f(0) = 0, so when x > 0, f(x) >0

i.e. when x > 0 (1+x) p > x p + 1, let x = a b,a,b > 0, then: (1+a b) p > a b) p + 1

Multiply b p on both sides to obtain: (a+b) p>=a p+b p

On the left, a p+a (p-1)*b+a (p-2)*b 2+.b^p

And p>1, p is a real number, a>0, b>0, so a (p-1)*b+a (p-2)*b 2+...a*b (p-1)>0, so a p+a (p-1)*b+a (p-2)*b 2+...b^p>a^p+b^p

i.e. (a+b) p>a p+b p

x~n(3,4)，∴x-3)/2~n(0,1)。

1)，p=p(x>2)+p(x<-2)。

and p(x>2)=p[(x-3) 2>(2-3) 2=-1 2]=1- (1 2)= 1 2); p(x<-2)=p[(x-3)/2<(-2-3)/2=-5/2]=φ5/2)=1-φ(5/2)。Check the standard normal distribution table to dig (1 2)=,p= (1 2)+1- (5 2)=.

2)，p=p[(x-3)/2>(3-3)/2=0]=1-φ(0)。and (0)=1 2, p=1-1 2=1 2.

If a is a regular set of b, then p(b-a) = p(b)-p(a).

If a and b do not intersect, then p(a b) = p(a) + p(b).

Using the above two items, p(a b) = p(a (b-(a b)) = p(a) + p(b-(a old next to b)) = p(a) + p(b)-p(a b).

From the lead of the world to p(a ant lack b) = p(a) + p(b)-p(ab) then search: p((a b) c) = p(a b) + p(c)-p((a b)c) = p(a) + p(b)-p(ab) + p(c)-p(ac bc) = p(a) + p(b) + p(c)-p(ab)-p(ac)-p(bc) + p(ac bc) = p(a) + p(b) + p(a) + p(b) - p(ab)-p(ac)-p(bc)+p(abc)

Evidence Lu type: the second unequal sign of the first certificate:

1 (1+x p)<1, both sides can be integrated at the same time.

Let's open the first inequality:

Notice the inequality with early guessing:

1/(1+x^p)>1-x^p

Integrating both sides at the same time gives the first inequality.

With b'Indicates the opposing event of b, then the wild thing p(b'Yinshen) = 1-p(b), a and b are mutually exclusive, and song pulse fluid a is established when b'Founded, ab'=a,∴p(ab') = p(a), by the conditional probability formula, p(a|b')=p(ab')/p(b')=p(a)/[1-p(b)].

Let f(x)=x p,x>0

From the hail 0, we may wish to let a(a+b) p-b p=f(a+b)-f(b) f(a+b)-f(b) from the Lagrangian mid-width return sail value theorem to obtain f(a+b)-f(b)=f'(c)(a+b-b)=f'(c)*a，b<=c<=a+b

a p=f(a)-f(0), the same gives f(a)-f(0)=f'(d)*a,0 "World stupidity=d<=a."

Decreasing by f(x), f'(c) i.e. (a+b) p<=a p+b p, and a or b is equal to 0.

A simple application of the differential median value theorem.

Let f(x)=x p,x>0

From the 0 untraceable, let a then (a+b) p-b p=f(a+b)-f(b) f(b) from the Lagrangian median Zi Xunshu theorem f(a+b)-f(b)=f'(c)(a+b-b)=f'(c)*a，b<=c<=a+b

a p=f(a)-f(0), the same gives f(a)-f(0)=f'(d)*a，0<=d<=a

Decreasing by f(x), f'(c) i.e. (a+b) p<=a p+b p, a or b is equal to changqiao 0.