Question

Answer 1

Ta có:

\(xy+xz=x\left(y+z\right)\le\dfrac{1}{4}\left(x+y+z\right)^2=\dfrac{1}{4}\) (đpcm)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\dfrac{1}{2};\dfrac{1}{4};\dfrac{1}{4}\right)\)

Answer 2

-Áp dụng BĐT AM-GM ta có:

\(xy+xz=x\left(y+z\right)\le\dfrac{\left[x+\left(y+z\right)\right]^2}{4}=\dfrac{1}{4}\)