Bài144: Cho các số thực dương x,y,z thỏa mãn điều kiện $x^{2}+y^{2}+z^{2}=1$.Tim min:
$P=4xy+4yz+10xz+\frac{1}{(x+z)(x+y+z)+1}$
Thực cũng ăn ngon
Ta có:
$P=2\left ( x+y+z \right )^{2}+3\left ( z+x \right )^{2}+\frac{1}{\left ( x+z \right )\left ( x+y+z \right )+1}+3y^{2}-5\\\geq 2\left ( x+y+z \right )^{2}+3\left ( x+z \right )^{2}++\frac{1}{\left ( x+z \right )\left ( x+y+z \right )+1}-5$
Đặt $\left\{\begin{matrix} x+y+z=a & \\ y+z=b & \end{matrix}\right.$
Khi đó: $P\geq 2a^{2}+3b^{2}+\frac{1}{ab+1}-5$
Ta sẽ chứng minh $ab+1> 0$. Thật vậy:
$ab+1=\left ( x+z \right )\left ( x+y+z \right )+x^{2}+y^{2}+z^{2}\\=\left ( x+z \right )^{2}+x^{2}+y^{2}+z^{2}+xy+yz\\\geq \frac{1}{2}\left ( x+z \right )^{2}+y\left ( x+z \right )+\frac{1}{2}y^{2}+\frac{1}{2}y^{2}\\=\frac{1}{2}\left ( x+y+z \right )^{2}+\frac{1}{2}y^{2}\geq 0$
Dấu "=" không xảy ra, ta có $ab+1> 0$
Vậy: $P\geq 2\sqrt{6}\left | ab \right |+\frac{1}{\left | ab \right |+1}-5\\\geq \left ( \left | ab \right |+1 \right )+\frac{1}{\left | ab \right |+1}-6\\\geq -4$
Vậy $\min P=-4\Leftrightarrow \left\{\begin{matrix} x=-z=-\frac{1}{\sqrt{2}} & \\ y=0 & \end{matrix}\right.\vee \left\{\begin{matrix} x=-z=\frac{1}{\sqrt{2}} & \\ y=0 & \end{matrix}\right.$