Lời giải:
Do \(x+y+z=1\) nên biến đổi như sau:
\(P=\frac{x}{(x+y)+(x+z)}+\frac{y}{(y+z)+(y+x)}+\frac{z}{(z+x)+(z+y)}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{(x+y)+(x+z)}\leq \frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\Rightarrow \frac{x}{(x+y)+(x+z)}\leq \frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\)
Thực hiện tương tự với các phân thức còn lại:
\(\Rightarrow P\leq \frac{1}{4}\left(\frac{x+y}{x+y}+\frac{y+z}{y+z}+\frac{z+x}{x+z}\right)=\frac{3}{4}\)
Vậy \(P_{\max}=\frac{3}{4}\Leftrightarrow x=y=z=\frac{1}{3}\)
\(P=\dfrac{x}{x+1}+\dfrac{y}{y+1}+\dfrac{z}{z+1}\)
Thay \(x+y+z=1\) vào biểu thức
\(\Rightarrow P=\dfrac{x}{2x+y+z}+\dfrac{y}{x+2y+z}+\dfrac{z}{x+y+2z}\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2x+y+z}=\dfrac{x}{x+y+x+z}\le\dfrac{x}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\\\dfrac{y}{x+2y+z}=\dfrac{y}{x+y+y+z}\le\dfrac{y}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}\right)\\\dfrac{z}{x+y+2z}=\dfrac{z}{x+z+y+z}\le\dfrac{z}{4}\left(\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{x}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)+\dfrac{y}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}\right)+\dfrac{z}{4}\left(\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)\)
\(\Rightarrow VT\le\dfrac{x}{4\left(x+y\right)}+\dfrac{x}{4\left(x+z\right)}+\dfrac{y}{4\left(x+y\right)}+\dfrac{y}{4\left(y+z\right)}+\dfrac{z}{4\left(x+z\right)}+\dfrac{z}{4\left(y+z\right)}\)
\(\Rightarrow VT\le\dfrac{x}{4\left(x+y\right)}+\dfrac{y}{4\left(x+y\right)}+\dfrac{x}{4\left(x+z\right)}+\dfrac{z}{4\left(x+z\right)}+\dfrac{y}{4\left(y+z\right)}+\dfrac{z}{4\left(y+z\right)}\)
\(\Rightarrow VT\le\dfrac{x+y}{4\left(x+y\right)}+\dfrac{x+z}{4\left(x+z\right)}+\dfrac{y+z}{4\left(y+z\right)}\)
\(\Rightarrow VT\le\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)
\(\Rightarrow P\le\dfrac{3}{4}\)
Vậy \(P_{max}=\dfrac{3}{4}\)
Dấu '' = '' xảy ra khi \(x=y=z\)
Áp dụng bđt Cauchy-Schwraz dạng Engel ta có:
\(P=\left(1-\dfrac{1}{x+1}\right)+\left(1-\dfrac{1}{y+1}\right)+\left(1-\dfrac{1}{z+1}\right)\)
\(=3-\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right)\le3-\dfrac{\left(1+1+1\right)^2}{x+1+y+1+z+1}\)
\(\le3-\dfrac{9}{4}=\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)