Đề nhảm.a;b;c ở đâu bạn -_-
a) Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel:
\(\left\{{}\begin{matrix}\dfrac{x}{2x+y+z}=\dfrac{x}{x+y+x+z}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)\\\dfrac{y}{2y+x+z}=\dfrac{y}{x+y+y+z}\le\dfrac{1}{4}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right)\\\dfrac{z}{2z+x+y}=\dfrac{z}{x+z+y+z}\le\dfrac{1}{4}\left(\dfrac{z}{x+z}+\dfrac{z}{y+z}\right)\end{matrix}\right.\)
Cộng theo vế:
\(\dfrac{x}{2x+y+z}+\dfrac{y}{2y+x+z}+\dfrac{z}{2z+x+y}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{y+z}+\dfrac{x}{x+z}+\dfrac{z}{x+z}\right)=\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(x=y=z>0\)
b) Áp dụng bất đẳng thức AM-GM:
\(\left\{{}\begin{matrix}\left(a+b-c\right)\left(a-b+c\right)\le\dfrac{\left(a+b-c+a-b+c\right)^2}{4}=\dfrac{4a^2}{4}=a^2\\\left(a-b+c\right)\left(-a+b+c\right)\le\dfrac{\left(a-b+c-a+b+c\right)^2}{4}=\dfrac{4c^2}{4}=c^2\\\left(a+b-c\right)\left(-a+b+c\right)\le\dfrac{\left(a+b-c-a+b+c\right)^2}{4}=\dfrac{4b^2}{4}=b^2\end{matrix}\right.\)
Nhân theo vế: \(\left[\left(a+b-c\right)\left(a-b+c\right)\left(-a+b+c\right)\right]^2\le\left(abc\right)^2\)
\(\Rightarrow\left(a+b-c\right)\left(a-b+c\right)\left(-a+b+c\right)\le abc\)
Dấu "=" xảy ra khi: \(a=b=c>0\)
