Another proof of APMO inequality

Aassila's inequality

a,b,c are positive reals
show that:

APMO inequality

If a,b,c>0 show that (a²+2)(b²+2)(c²+2)≥9(ab+bc+ca)

Prove: Let          
we just need to show that



in fact we can easily got:








Now the original inequality is equivalent to
2(xy+yz+zx)-2(xy²+yz²+zx²)-2(x²y+y²z+z²x)≤2/9 
xy²+yz²+zx²+x²y+y²z+z²x+1/9≥xy+yz+zx


by AM-GM we obtain:


                                                           

                                                                                                              Q.E.D

Triangle inequality

Let  then we have:

  Prove:

We can use make this geometry graph

Let A_iB_i+1=a_i

C_iC_i+1=b_i

by Pythagoras's theorem we have:

because the shortest distance of two point is a line segment

we obviously have:

       
the equality holds if and only if                                                 Q.E.D

Let a,b,c be the positive reals
Show that



we can apply triangle inequality:


The inequality is now equivalent to

 where x=a+b+c,but this is the easy exercise of quadratic function.

Let ABCD be a quadrilateral E,F,G,H be the middle points of AB BC CD DA

Prove that

some Geometric inequalities

Let a,b,c be the lengths of a triangle R be the radius of the circumcircle
p=a+b+c s=(a+b+c)/2
there we have:

An easy one

forall a,b,c>0
 we have



see here:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=466791&p=2619149#p2619149

A tip.

In order to prove ∑F(x,y,z)>=C,we can consider that to prove

F(x,y,z)>=G(x,y,z) where ∑G(x,y,z)=C

inequality

Let a,c,b>0 a+b+c=2

prove: