TA’s Challenge Problem #3

**TheAbstractionist** · June 9th 2009, 02:18 AM

If $x,y,z$ are non-negative real numbers, show that

$(x+y+z)^2+xy+yz+zx\ \geqslant\ 2\sqrt{xy}(x+y)+2\sqrt{yz}(y+z)+2\sqrt{zx}(z+x)$

**NonCommAlg** · June 9th 2009, 03:36 PM

Originally Posted by TheAbstractionist

If $x,y,z$ are non-negative real numbers, show that

$(x+y+z)^2+xy+yz+zx\ \geqslant\ 2\sqrt{xy}(x+y)+2\sqrt{yz}(y+z)+2\sqrt{zx}(z+x)$

we have $(x+y-2\sqrt{xy})^2 \geq 0,$ which gives us: $x^2 + y^2+6xy \geq 4\sqrt{xy}(x+y).$ similarly: $y^2 + z^2+6yz \geq 4\sqrt{yz}(y+z)$ and $x^2 + z^2+6xz \geq 4\sqrt{xz}(x+z).$

add these three inequalities together and then divide by 2 to get the result.

**TheAbstractionist** · June 9th 2009, 11:54 PM

Very good!

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