\(P=\dfrac{x^3+y^3+z^3}{xy+2yz+zx}=\dfrac{x^3}{xy+2yz+zx}+\dfrac{y^3}{xy+2yz+zx}+\dfrac{z^3}{xy+2yz+zx}\)\(\ge\sqrt[3]{\dfrac{x^3\cdot y^3\cdot z^3}{\left(xy+2yz+zx\right)^3}}=\dfrac{xyz}{xy+2yz+zx}\)

ta có: (x+y+z)^2≥0 <=>xy+yz+zx ≥\(-\dfrac{x^2+y^2+z^2}{2}\) (1)

(y+z)^2 ≥ 0 <=> yz ≥ \(-\dfrac{y^2+z^2}{2}\) (2)

(1), (2) => xy+2yz+zx ≥ \(-\dfrac{x^2}{2}\)

-.-

trên tử là x^2+y^2+z^2 nhé bài này dùng pp điểm rơi giả định :v

Thấy cái đề mà thấy khiếp ...

Ta có : \(x^2-xy+y^2=\frac{3}{4}\left(x^2-2xy+y^2\right)+\frac{1}{4}\left(x^2+2xy+y^2\right)\)

                                       \(=\frac{3}{4}\left(x-y\right)^2+\frac{1}{4}\left(x+y\right)^2\ge\frac{1}{4}\left(x+y\right)^2\)

\(\Rightarrow\sqrt{x^2-xy+y^2}\ge\frac{x+y}{2}\)

Tương tự \(\sqrt{y^2-yz+z^2}\ge\frac{y+z}{2}\)

                \(\sqrt{z^2-zx+x^2}\ge\frac{x+z}{2}\)

Do đó : \(2S\ge\frac{x+y}{x+y+2z}+\frac{y+z}{y+z+2x}+\frac{x+z}{x+z+2y}\)

\(\Rightarrow2S+3\ge\left(1+\frac{x+y}{x+y+2z}\right)+\left(1+\frac{y+z}{y+z+2x}\right)+\left(1+\frac{x+z}{x+z+2y}\right)\)

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

                                                         \(\ge2\left(x+y+z\right).\frac{9}{4\left(x+y+z\right)}\)\(=\frac{9}{2}\)

                                                          (Áp dụng bđt Cô-si dạng engel cho 3 số)

\(\Rightarrow2S+3\ge\frac{9}{2}\)

\(\Rightarrow S\ge\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)

Vậy ..............

Cho xin cái k đúng cái

Xét:

\(P-\left(\sqrt{3}-1\right)=\frac{1}{4}\left[2x-\left(\sqrt{3}-1\right)\left(y+z\right)\right]^2+\frac{\sqrt{3}}{2}\left(y-z\right)^2\ge0\)

\(\Rightarrow P\ge\sqrt{3}-1\)

Còn về dấu "=" xảy ra thì bạn có thể tự làm nhé :D

Mình ghi thiếu \(xy+2yz+zx\) ở dưới mẫu nhé.

\(\frac{\left(x+y+z\right)^2}{3}\ge xy+yz+zx\Rightarrow x+y+z\ge3\)

\(P=\frac{x^2}{\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}}+\frac{y^2}{\sqrt{\left(y+2\right)\left(y^2-2y+4\right)}}+\frac{z^2}{\sqrt{\left(z+2\right)\left(z^2-2z+4\right)}}\) 

\(\Rightarrow P\ge\frac{\left(x+y+z\right)^2}{\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}+\sqrt{\left(y+2\right)\left(y^2-2y+4\right)}+\sqrt{\left(z+2\right)\left(z^2-2z+4\right)}}\)  

\(\Rightarrow P\ge\frac{2\left(x+y+z\right)^2}{\left(x+2+x^2-2x+4\right)+\left(y+2+y^2-2y+4\right)+\left(z+2+z^2-2z+4\right)}\) 

\(\Rightarrow P\ge\frac{2\left(x+y+z\right)^2}{\left(x^2+y^2+z^2\right)-\left(x+y+z\right)+18}=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-\left(x+y+z\right)-2\left(xy+yz+zx\right)+18}=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-\left(x+y+z\right)+12}\)

Dự đoán Min P=1 khi x+y+z=3

Đặt \(t=x+y+z\ge3\) 

\(\Rightarrow P\ge\frac{2t^2}{t^2-t+12}\Rightarrow P-1\ge\frac{t^2+t-12}{t^2-t+12}=\frac{\left(t-3\right)\left(t+4\right)}{t^2-t+12}\ge0\) 

\(\Rightarrow P\ge1\)

bạn là một thiên tài