\(\dfrac{x^2}{y+z}+\dfrac{y+z}{4}\ge x;\dfrac{y^2}{z+x}+\dfrac{z+x}{4}\ge y;\dfrac{z^2}{x+y}+\dfrac{x+y}{4}\ge z\)
\(\Rightarrow P\ge x+y+x-\dfrac{x+y+z}{2}=\dfrac{4}{2}=2\)
Vậy, \(GTNN\) của \(P=4\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(P=\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}=\dfrac{4}{2}=2\)
Đẳng thức xảy ra khi \(x=y=z=\dfrac{4}{3}\)
\(\dfrac{x^2}{y+z}+\dfrac{y+z}{4}\ge x;\dfrac{y^2}{z+x}+\dfrac{z+x}{4}\ge y;\dfrac{z^2}{x+y}+\dfrac{x+y}{4}\ge z\)
\(\Rightarrow P\ge x+y+x-\dfrac{x+y+z}{2}=\dfrac{4}{2}=2\)
Vậy, \(GTNN\) của \(P=2\)
P=\(\dfrac{x^2}{y+z}\)+\(\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\)
P+x+y+z=\(\dfrac{x^2}{y+z}+x+\dfrac{y^2}{x+z}+y+\dfrac{z^2}{x+y}+z\)
=\(\dfrac{x\left(x+y+z\right)}{y+z}+\dfrac{y\left(x+y+z\right)}{x+z}+\dfrac{z\left(x+y+z\right)}{x+y}\)
=(x+y+z)(\(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\))(*)
ta chứng minh bất đẳng thức phụ:
\(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\ge\dfrac{3}{2}\)
ta có:\(\dfrac{x}{y+z}+1+\dfrac{y}{x+z}+1+\dfrac{z}{x+y}+1\ge\dfrac{9}{2}\)
(x+y+z)(\(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\))\(\ge\dfrac{9}{2}\)
đặt a=x+y;b=y+z;c=z+x, ta có bất phương trình sau:
\(\dfrac{a+b+c}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{9}{2}\)(**)
mà a+b+c\(\ge3\sqrt[3]{abc}\);\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)
<=>(a+b+c)(\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\))\(\ge\)9
=>(**) được chứng minh
thay vào (*) ta được:P=(x+y+z)(\(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}-1\))
\(\ge\)4(\(\dfrac{3}{2}-1\))=2
cách khác nè
áp dụng BĐT svac :
P\(\ge\dfrac{\left(x+y+z\right)^2}{y+z+z+x+x+y}=\dfrac{16}{2\left(x+y+z\right)}=\dfrac{16}{8}=2\)
vậy Min P = 2
hic nhìn lên thấy ông ace legona làm zùi @@
