Inequality - prove that a^2/b + b^2/c + c^2/d + d^2/a > 4

[Guest]

a,b,c,d are positive real numbers such that [tex]a^2 + b^2 + c^2 + d^2 = 4[/tex]
Prove that [tex]\frac{a^2}{b } + \frac{b^2}{c } + \frac{c^2}{d } + \frac{d^2}{a } \ge 4[/tex]

[FarRider]

Sketch of proof using Holder http://watchmath.com/vlog/?p=185 and its inverse (when p,q<1)

Using GM/AM on [tex]a^2+b^2+c^2+d^2=4[/tex] we obtain [tex]abcd<=1[/tex].
Below we use cyclical sum notation:

(1) [tex]\sum_{c}^.{\frac{a^2}{b } }=\sum_{c}^.{\frac{a^4}{a^2b } }>=\frac{(\sum_{c}^.{a^2 } )^2}{\sum_{c}^.{a^2b } }=\frac{16}{\sum_{c}^.{a^2b } }[/tex]

This is obtained by (inverse) Holder with p=1/2 and q=-1.
Now we examine the denominator and apply Holder with p=q=1/2 for the tuples (a,b,c,d) and (ab,bc,cd,da).
(2) [tex]\sum_{c}^.{a^2b }<=\sqrt{\sum_{c}^.{ a^2} } \sqrt{\sum_{c}^.{ a^2b^2} }=2\sqrt{\sum_{c}^.{ a^2b^2} }<=2\sqrt{4abcd}<=4[/tex]

The second inequality above is GM-AM and the last follows from our initial observation that abcd<=1.

Plugging in (2) into (1) we obtain the result.

[Guest]

[tex]2\sqrt{\sum_{c}^.{ a^2b^2} }<=2\sqrt{4abcd}[/tex]

AM-GM goes in the other direction i.e ∑(a²b²) ≥ 4abcd

[FarRider]

Sorry, of course you are right. here is how it should finish:

[tex]\sum_{c}^.{ a^2b^2}=(a^2+c^2)(b^2+d^2)=(a^2+c^2)(4-a^2-c^2)<=4[/tex]

[FarRider]

In fact the same proof gives a general result:

[tex]\frac{a^2}{b } + \frac{b^2}{c } + \frac{c^2}{d } + \frac{d^2}{a }>=2\sqrt{a^2+b^2+c^2+d^2}[/tex]