Proof of inequalities. Find out more about the process

On both sides, the original inequality is equivalent to:

a1b2)²+a1b3)² a2b3)²+a2b1)²+a3b1 )²a3b2)²≥2a1b1a2b2+2a1b1a3b3+2a2b2a3b3

That is, the evidence: a1b2) -2a1b1a2b2+(a2b1) a3b1 ) 2a1b1a3b3+(a1b3) a2b3) 2a2b2a3b3+(a3b2) 0

That is: a1b2- a2b1) a3b1 -a1b3) a2b3-a3b2) 0

Apparently true. The above steps are reversible, and the original proposition is true.

Proof: 00

then x m-1 m (x-1).

Let x=a b, then (a b) m-1 m(a b-1), i.e., a m·b (-m) m(a b-1)+1

Multiply b on both sides of the inequality at the same time, yielding: a m·b (1-m) m(a-b)+b=ma+(1-m)b

Let m=1 p, a=a p, b=b q, then 1-m=1-1 p=1 q be substituted into a m·b (1-m) ma+(1-m)b, and we get:

ab≤a^p/p+b^q/q

Let ab=x, bc=y, ca=z

then the original inequality is equivalent to:

x^2+y^2+z^2>=xy+yz+zx=>2(x^2+y^2+z^2)>=2(xy+yz+zx)=>(x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)>=0

(x-y) 2+(y-z) 2+(z-x) 2>=0 is clearly true.

The constructor f(x)=x n(x>0,n is a positive integer), let a(x,x n),b(y,y n) be any two points on the function image, then the midpoint c of the line segment ab is ((x+y) 2,(x n+y n) 2), let d((x+y) 2,((x+y) 2) n), according to the image of the function f(x), we can know that the point c is above the point d, if and only if a,b coincides with the point c and the point d, so (x n+y n) 2 ((x+y) 2) n is x n+y n a n 2 (n-1)

in the holder inequality.

u1+v1)(u2+v2)..un+vn)>=(u1u2...un+v1v2...vn)^n

In the middle of the order u1=x, v1=y, all other parameters are 1, stand.