Áp dụng BĐT cosi ta có:
`x^{2}+yz>=2xsqrt{yz}`
`=>1/(x^{2}+yz)<=1/(2xsqrt{yz})`
Hoàn toàn tương tự:
`1/(y^{2}+zx)<=1/(2ysqrt{zx})`
`1/(z^{2}+xy)<=1/(2zsqrt{xy})`
`=>1/(x^{2}+yz)+1/(y^{2}+zx)+1/(z^{2}+xy)<=1/2(1/(xsqrt{yz})+1/(ysqrt{zx})+1/(zsqrt{xy}))`
`<=>1/(x^{2}+yz)+1/(y^{2}+zx)+1/(z^{2}+xy)<=1/2((sqrt{xy}+sqrt{yz}+sqrt{zx})/(xyz))(1)`
Đặt `sqrtx=a,sqrty=b,sqrtz=c`
Dễ thấy:
`a^{2}+b^{2}+c^{2}>=ab+bc+ca`
`=>sqrt{xy}+sqrt{yz}+sqrt{zx}<=x+y+z`
`=>1/2((sqrt{xy}+sqrt{yz}+sqrt{zx})/(xyz))<=1/2((x+y+z)/(xyz))`
`<=>1/2((sqrt{xy}+sqrt{yz}+sqrt{zx})/(xyz))<=1/2(1/(xy)+1/(yz)+1/(zx))(2)`
Từ `(1),(2)=>1/(x^{2}+yz)+1/(y^{2}+zx)+1/(z^{2}+xy)<=1/2(1/(xy)+1/(yz)+1/(zx))`
Dấu "=" xảy ra khi `x=y=z`
