`CM:1/(x^2+yz)+1/(y^2+zx)+1/(z^2+xy)<1/2( (1/(xy)+1/(yz)+1/(zx))`

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`