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simpler counterexample
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Peter Mueller
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This inequality doesn't hold in general. It is false for instance for $m=8$$m=7$, $n=9$$n=8$: Set $X=7/10$$X=3/4$, $Y=1$, and $Z=23/17$$Z=9/7$. Then $XY+YZ+ZX=3$, however $\frac{X^9}{X^8+Y^8}+\frac{Y^9}{Y^8+Z^8}+\frac{Z^9}{Z^8+X^8}<\frac{3}{2}$$\frac{X^8}{X^7+Y^7}+\frac{Y^8}{Y^7+Z^7}+\frac{Z^8}{Z^7+X^7}<\frac{3}{2}$.

This inequality doesn't hold in general. It is false for instance for $m=8$, $n=9$: Set $X=7/10$, $Y=1$, and $Z=23/17$. Then $XY+YZ+ZX=3$, however $\frac{X^9}{X^8+Y^8}+\frac{Y^9}{Y^8+Z^8}+\frac{Z^9}{Z^8+X^8}<\frac{3}{2}$.

This inequality doesn't hold in general. It is false for instance for $m=7$, $n=8$: Set $X=3/4$, $Y=1$, and $Z=9/7$. Then $XY+YZ+ZX=3$, however $\frac{X^8}{X^7+Y^7}+\frac{Y^8}{Y^7+Z^7}+\frac{Z^8}{Z^7+X^7}<\frac{3}{2}$.

Source Link
Peter Mueller
  • 22.5k
  • 1
  • 75
  • 107

This inequality doesn't hold in general. It is false for instance for $m=8$, $n=9$: Set $X=7/10$, $Y=1$, and $Z=23/17$. Then $XY+YZ+ZX=3$, however $\frac{X^9}{X^8+Y^8}+\frac{Y^9}{Y^8+Z^8}+\frac{Z^9}{Z^8+X^8}<\frac{3}{2}$.