áp dụng bdt cô-si
\(\sqrt{\frac{y+z}{x}\cdot1}\le\left(\frac{y+z}{x}+1\right):2=\frac{x+y+z}{2x}\)
\(\Rightarrow\sqrt{\frac{x}{y+z}}\ge\frac{2x}{x+y+z}\)
bạn chứng minh tương tự ta cx có
\(\sqrt{\frac{y}{x+z}}\ge\frac{2y}{x+y+z};\sqrt{\frac{z}{y+x}}\ge\frac{2z}{x+y+z}\)
cộng từng vế lại vs nhau ta có \(\sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{x+z}}+\sqrt{\frac{z}{x+y}}\ge\frac{2\left(x+y+z\right)}{x+y+z}=2\)
dấu = xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x=y+z\\y=z+x\\z=x+y\end{cases}}\Rightarrow x+y+z=0\ne gt\)
suy ra đẳng thức ko xảy ra
sau khi đc hồn ông cô si nhập vô,mình đã làm ra
ta có:
\(\left(x+y+z\right)^2\ge4x\left(y+z\right);\left(x+y+z\right)^2\ge4y\left(x+z\right);\left(x+y+z\right)^2\ge4z\left(x+y\right)\)
\(\Leftrightarrow\frac{\left(x+y+z\right)^2}{4x}\ge y+z;\frac{\left(x+y+z\right)^2}{4y}\ge z+x;\frac{\left(x+y+z\right)^2}{4z}\ge x+y\)
\(\Rightarrow\sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{z+x}}+\sqrt{\frac{z}{x+y}}\ge\sqrt{\frac{4x^2}{\left(x+y+z\right)^2}}+\sqrt{\frac{4y^2}{\left(x+y+z\right)^2}}+\sqrt{\frac{4z^2}{\left(x+y+z\right)^2}}\)
\(=\frac{2x}{x+y+z}+\frac{2y}{x+y+z}+\frac{2z}{x+y+z}=2\)
đẳng thức xảy ra khi \(\hept{\begin{cases}x=y+z\\y=z+x\\z=x+y\end{cases}\Leftrightarrow x=y=z=0}\)(vô lí)
\(\Rightarrow\sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{z+x}}+\sqrt{\frac{z}{x+y}}>2\left(Q.E.D\right)\)
