[tatari/nightmare]

พิจารณา
$\frac{x^2-yz}{x^2+2yz}-1=\frac{-3yz}{x^2+2yz}$
$$\therefore \frac{x^2-yz}{x^2+2yz}+\frac{y^2-zx}{y^2+2zx}+\frac{z^2-xy}{z^2+2xy}-3+3$$
$$=3+(\frac{x^2-yz}{x^2+2yz}-1)+(\frac{y^2-zx}{y^2+2zx}-1)+(\frac{z^2-xy}{z^2+2xy}-1)$$
$$=3+\frac{-3yz}{x^2+2yz}+\frac{-3zx}{y^2+2zx}+\frac{-3xy}{z^2+2xy}$$
$$=3-3[\frac{yz}{x^2+2yz}+\frac{zx}{y^2+2zx}+\frac{xy}{z^2+2xy}]$$
ต่อไปจะแสดงว่า
$$\frac{yz}{x^2+2yz}+\frac{zx}{y^2+2zx}+\frac{xy}{z^2+2xy}\leq 1$$
ซึ่งจะเห็นว่า $\frac{yz}{x^2+2yz}\leq 1$,$\frac{zx}{y^2+2zx}\leq1$,$\frac{xy}{z^2+2xy}\leq1$