Cho x,y,z >0 . Chứng minh:
[tex]\frac{x^{2}}{(x+y)(x+z)}+\frac{y^{2}}{(x+y)(y+z)}+\frac{z^{2}}{(x+z)(y+z)}\geq \frac{3}{4}[/tex]
Theo BĐT Cauchy ta có:
[tex]x^2y+x^2z+y^2z+y^2x+z^2x+z^2y\geq 6\sqrt[6]{x^2y.x^2z.y^2z.y^2x.z^2x.z^2y}=6xyz\\\Leftrightarrow 4(x^2y+x^2z+y^2z+y^2x+z^2x+z^2y)\geq 6xyz+3(x^2y+x^2z+y^2z+y^2x+z^2x+z^2y)\\\Leftrightarrow 4[x^2(y+z)+y^2(z+x)+z^2(x+y)]\geq 3(2xyz+x^2y+xz^2+x^2z+y^2z+y^2x+yz^2)\\\Leftrightarrow 4[x^2(y+z)+y^2(z+x)+z^2(x+y)]\geq 3(x+y)(y+z)(z+x)\\\Leftrightarrow \frac{x^2(y+z)+y^2(z+x)+z^2(x+y)}{(x+y)(y+z)(z+x)}\geq \frac{3}{4}(vì:x,y,z>0)\\\Leftrightarrow \frac{x^{2}}{(x+y)(x+z)}+\frac{y^{2}}{(x+y)(y+z)}+\frac{z^{2}}{(x+z)(y+z)}\geq \frac{3}{4} [/tex]
Dấu = xảy ra khi [tex]x=y=z[/tex]