Question

Cho x;y;z > 0 Tìm Min.

P = \(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2zx}+\frac{z^2}{z^2+2xy}\)

Answer 1

Vs x,y,z>0 .Áp dụng bđt Svac-xơ có:

\(P=\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}\ge\frac{\left(x+y+z\right)^2}{x^2+2yz+y^2+2xz+z^2+2xy}\)

<=> P\(\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}\)

<=> P\(\ge1\)

Dấu "=" xảy ra<=> x=y=z=1

Vậy minP=1 <=> x=y=z=1

Answer 2

Solution:

Áp dụng BĐT Cauchy-Schwarz :

\(P\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2xy+2yz+2zx}=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)