dùng co si Swartch nhé bạn
TQ :\(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\)
\(\frac{x^2}{y+z}+\frac{y+z}{4}\ge2\sqrt{\frac{x^2\left(y+z\right)}{4\left(y+z\right)}}\Rightarrow\frac{x^2}{y+z}+\frac{y+z}{4}\ge x\)(BĐT Cô-si)
Tương tự \(\Rightarrow\frac{y^2}{z+x}+\frac{z+x}{4}\ge y\)
\(\frac{z^2}{x+y}+\frac{x+y}{4}\ge z\)
Cộng vế với vế lại\(\Rightarrow\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}+\frac{x+y+z}{2}\ge x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\ge\frac{x+y+z}{2}=1\)
Dấu "=" xảy ra <=> x=y=z=2/3
