ta có: \(\frac{x^2}{y+z}+\frac{y+z}{4}\ge2\sqrt{\frac{x^2}{y+z}.\frac{y+z}{4}}=x\)(dấu = xảy ra khi \(\left(y+z\right)^2=4x^2\)↔y+z=2x)

tương tự ta có:\(\frac{y^2}{x+z}+\frac{x+z}{4}\ge y;\frac{z^2}{x+y}+\frac{x+y}{4}\ge z\)(dấu = cũng xảy ra khi x+z=2y;x+y=2z)

cộng từng vế ta có:P+\(\frac{x+y+z}{2}\ge x+y+z\)

→P\(\ge\frac{x+y+z}{2}\)mà x+y+x=1

\(P\ge\frac{1}{2}\)↔\(\begin{cases}y+z=2x\\x+z=2y\\x+y=2z\end{cases}\)→x=y=z=1/3

\(\frac{x}{1+y^2}=x-\frac{xy^2}{1+y^2}\ge x-\frac{xy^2}{2y}=x-\frac{1}{2}xy\)

Tương tự và cộng lại:

\(A\ge x+y+z-\frac{1}{2}\left(xy+yz+zx\right)\ge x+y+z-\frac{1}{6}\left(x+y+z\right)^2=\frac{3}{2}\)

\("="\Leftrightarrow x=y=z=1\)

Áp dụng Bunhia.

\(\left(x+y+z\right)^2\le\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)=3.3=9\)

=> \(0< x+y+z\le3\)

Có: \(P=\frac{x^2+1}{x}+\frac{y^2+1}{y}+\frac{z^2+1}{z}-\frac{1}{x+y+z}\)

\(=\frac{x^2-2x+1}{x}+\frac{y^2-2y+1}{y}+\frac{z^2-2z+1}{z}-\frac{1}{x+y+z}+6\)

\(=\frac{\left(x-1\right)^2}{x}+\frac{\left(y-1\right)^2}{y}+\frac{\left(z-1\right)^2}{z}-\frac{1}{x+y+z}+6\)

\(\ge\frac{\left(x+y+z-3\right)^2}{x+y+z}-\frac{1}{x+y+z}+6=\frac{\left(x+y+z-3\right)^2-1}{x+y+z}+6\)

\(\ge\frac{0-1}{3}+6=\frac{17}{3}\)

"=" xảy ra <=> \(x+y+z=3;x=y=z\Leftrightarrow x=y=z=1\)

Vậy min P = 17/3 <=> x = y = z =1.

\(P=\frac{x^2+1}{x}+\frac{y^2+1}{y}+\frac{z^2+1}{z}-\frac{1}{x+y+z}\)

\(=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}\)

\(\ge x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(=x+y+z+\frac{8x}{9}+\frac{8y}{9}+\frac{8z}{9}\)

Có BĐT phụ \(a+\frac{8}{9a}\ge\frac{a^2+33}{18}\)

\(\Leftrightarrow\frac{9a^2+8}{9a}\ge\frac{a^2+33}{18}\)

\(\Leftrightarrow162a^2+144-9a^3-297a\ge0\)

\(\Leftrightarrow-a^3+18a^2-33a+16\ge0\)

\(\Leftrightarrow\left(a-1\right)^2\left(16-a\right)\ge0\left(OK\right)\)

\(\Rightarrow P\ge\frac{x^2+y^2+z^2+99}{18}=\frac{17}{3}\)

Dấu "=" xảy ra tại x=y=z=1

\(\frac{17}{3}\) đúng k?

Áp dụng bất đẳng thức cosi schwarz 

 \(A\ge\frac{\left(x+y+z\right)^2}{2x^2+2y^2+2z^2+5\left(xy+yz+xz\right)}=\frac{9}{18+\left(xy+yz+xz\right)}\)

Mà \(xy+yz+xz\le\frac{1}{3}\left(x+y+z\right)^2=3\)

=> \(A\ge\frac{9}{18+3}=\frac{3}{7}\)

MinA=3/7

Dấu bằng xảy ra khi x=y=z=1