xyzx2y2+y2z2+z2x2≥3
(x2y2+y2z2+z2x2)2≥3x2y2z2
(x4y4+y4z4+z4x4)+2(x4y2z2+x2y4z2+x2y2z4)≥3x2y2z2
(x4y4+y4z4+z4x4)+2(x2+y2+z2)x2y2z2≥3x2y2z2
(x4y4+y4z4+z4x4)+2x2y2z2≥3x2y2z2
(x4y4+y4z4+z4x4)≥x2y2z2
falls;x2=y2=z2=31
3(31)2(31)2=271=31∗31∗31
x2−a=y2−b=z2−c=31
1≥a;b;c≥−1
a+b+c=0
(31+a)2(31+b)2+(31+b)2(31+c)2+(31+c)2(31+a)2≥(31+a)∗(31+b)∗(31+c)
(91+32a+a2)(91+32b+b2)+(91+32b+b2)+(91+32c+c2)(91+32a+a2)≥271+91(a+b+c)+31(ab+bc+ca)+abc
(811+272(a+b)+91(a2+b2)+94(ab)+32(a2b+ab2)+a2b2)+(811+272(b+c)+91(b2+c2)+94(bc)+32(b2c+bc2)+b2c2)+(811+272(a+c)+91(a2+c2)+94(ac)+32(a2c+ac2)+a2c2)≥271+91(a+b+c)+31(ab+bc+ca)+abc
(91(a2+b2)+94(ab)+32(a2b+ab2)+a2b2)+(91(b2+c2)+94(bc)+32(b2c+bc2)+b2c2)+(91(a2+c2)+94(ac)+32(a2c+ac2)+a2c2)≥31(ab+bc+ca)+abc
(32(a2+b2+c2)+94(ab+bc+ca)+32(a2(b+c)+b2(c+a)+c2(a+b))+(a2b2+b2c2+c2a2)≥31(ab+bc+ca)+abc
(32(a2+b2+c2)+91(ab+bc+ca)+32(a2(b+c)+b2(c+a)+c2(a+b))+(a2b2+b2c2+c2a2)≥abc
Eine Idee ist c durch - (a+b) zu ersetzen.
1≥a≥b≥c≥−1
c=−(a+b)
(34(a2+ab+b2)−91(a2+ab+b2)+2(a2b+ab2)+(a2b2+(b2+a2)(a+b)2)≥−a2b−ab2
(311(a2+ab+b2)+3(a2b+ab2)+(a2b2+(b2+a2)(a+b)2)≥0ä
Stimmt für
1≥a≥b≥0≥c≥−1
Wenn
1≥a≥0≥b≥c≥−1
So ist
1∣a∣≥∣b∣
(311(a+b)2−ab)+3(a2b+ab2)+(a2b2+(b2+a2)(a+b)2)≥0ä
Wenn b<0, dann
311(−ab)+3a2b>0
Damit ist die Behauptung bewiesen.