Dễ dàng chứng minh được:

\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) với \(a,b,c>0\)(1)

Dấu bằng xảy ra \(\Leftrightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)

Theo đề bài, vì x, y, z > 0 nên áp dụng (1), ta có:

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

Vì x y, z > 0 nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được:

\(x+y\ge2\sqrt{xy}\)(3)

Chứng mih tương tự, ta được;

\(y+z\ge2\sqrt{yz}\)(4);

\(z+x\ge2\sqrt{zx}\)(5)

Từ (3), (4), (5), ta được:

\(2\left(x+y+z\right)\ge2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)

\(\Leftrightarrow x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

\(\Leftrightarrow2\left(x+y+z\right)\ge x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

\(\Leftrightarrow\frac{1}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\ge\)\(\frac{1}{2\left(x+y+z\right)}\)

\(\Leftrightarrow\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\ge\frac{x+y+z}{2}\)

Mà theo đề bài, \(x+y+z\ge3\) nên:

\(\frac{x+y+z}{2}\ge\frac{3}{2}\)

Suy ra \(\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\ge\frac{3}{2}\left(6\right)\)

Từ (2) và (6), ta được:

\(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{zx}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{3}{2}\)(điều phải chứng minh)

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}x=y=z\\x+y+z=3\end{cases}\Leftrightarrow x=y=z=1}\)

Vậy nếu x, y, z > 0 và \(x+y+z\ge3\)thì \(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{zx}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{3}{2}\)

\(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{xz}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{3}{2}\)

\(\frac{x^2}{x+\sqrt{yz}}+\frac{x+\sqrt{yz}}{4}\ge2\sqrt{\frac{x^2}{4}}=x\)

Tượng tự ta có : \(\frac{y^2}{y+\sqrt{xz}}+\frac{y+\sqrt{xz}}{4}\ge y\)

\(\frac{z^2}{z+\sqrt{xy}}+\frac{z+\sqrt{xy}}{4}\ge z\)

Cộng vế với vế của BĐT ta được : 

\(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{xz}}+\frac{z^2}{z+\sqrt{xy}}+\frac{x+\sqrt{yz}}{4}+\frac{y+\sqrt{xz}}{4}+\frac{z+\sqrt{xy}}{4}\ge x+y+z\)

\(VT\ge x+y+z-\frac{x+y+z+\sqrt{yz}+\sqrt{xz}+\sqrt{xy}}{4}\)

mà \(\sqrt{yz}+\sqrt{xz}+\sqrt{xy}\le x+y+z\)

\(VT\ge\frac{4\left(x+y+z\right)-2\left(x+y+z\right)}{4}=\frac{2\left(x+y+z\right)}{4}\)

mà \(x+y+z\ge3\)hay \(VT\ge=\frac{6}{4}=\frac{3}{2}\)

Dấu ''='' xảy ra <=> x = y = z = 1

Ta có: \(x+y+z=0\)

\(\Rightarrow x+y=-z\)

\(\Leftrightarrow\left(x+y\right)^2=\left(-z\right)^2\)

\(\Leftrightarrow x^2+2xy+y^2=z^2\)

\(\Leftrightarrow x^2+y^2-z^2=-2xy\)

Chứng minh tương tự ta có:

\(x^2+z^2-y^2=-2xz\)

\(y^2+z^2-x^2=-2yz\)

\(\frac{xy}{x^2+y^2-z^2}+\frac{xz}{x^2+z^2-y^2}+\frac{yz}{y^2+z^2-x^2}\)

\(=\frac{xy}{-2xy}+\frac{xz}{-2xz}+\frac{yz}{-2yz}\)

\(=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\)

\(=-\frac{3}{2}\)

Vậy giá trị biểu thức là \(-\frac{3}{2}\)

http://diendantoanhoc.net/topic/160455-%C4%91%E1%BB%81-to%C3%A1n-v%C3%B2ng-2-tuy%E1%BB%83n-sinh-10-chuy%C3%AAn-b%C3%ACnh-thu%E1%BA%ADn-2016-2017/

bài của tui mà -_-

hihi k biết làm nên đăng ^^

ta có:

\(S\ge\frac{x^3}{x^2+y^2+\frac{x^2+y^2}{2}}+\frac{y^3}{y^2+z^2+\frac{y^2+z^2}{2}}+\frac{z^3}{z^2+x^2+\frac{z^2+x^2}{2}}\)

\(\Rightarrow S\ge\frac{2x^3}{3\left(x^2+y^2\right)}+\frac{2y^3}{3\left(y^2+z^2\right)}+\frac{2z^3}{3\left(z^2+x^2\right)}\Rightarrow\frac{3}{2}S\ge P=\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2}\)

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

\(=\left(x+y+z\right)-\frac{1}{2}\left(x+y+z\right)=\frac{9}{2}\)

\(\Rightarrow\frac{3}{2}S\ge\frac{9}{2}\Rightarrow S\ge3\)

Vậy Min S=3 khi x=y=z=3

hok lp 6 000000000000 biet toan lp 9 dau ma lm , tk di , giai cho

\(VT=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\)

\(\ge\frac{3x}{y+z+1}+\frac{3y}{x+z+1}+\frac{3z}{x+y+1}\)

\(=\frac{3x^2}{xy+xz+x}+\frac{3y^2}{xy+yz+y}+\frac{3z^2}{xz+yz+z}\)

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

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

\(\ge\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\ge xy+yz+xz=VP\)

Dấu "=" <=> x=y=z=1

Bài 3 thì \(\le1\)

Bài 4 thì \(\ge\frac{3}{4}\) nhé