Ta có : \(\frac{x}{x^2-yz+2010}+\frac{y}{y^2-xz+2010}+\frac{z}{z^2-xy+2010}\)

\(=\frac{x^2}{x^3-xyz+2010x}+\frac{y^2}{y^3-xyz+2010y}+\frac{z^2}{z^3-xyz+2010z}\)

\(\ge\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2010\left(x+y+z\right)}=\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+3\left(xy+yz+xz\right)\left(x+y+z\right)}\)

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

vì x+y+z=1nên

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

nen \(\frac{xy}{x^2+y^2}+\frac{yz}{y^2+z^2}+\frac{xz}{x^2+z^2}+\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\) =\(\left(\frac{xy}{x^2+y^2}+\frac{x^2+y^2}{4xy}\right)+\left(\frac{yz}{y^2+z^2}+\frac{y^2+z^2}{4yz}\right)+\left(\frac{xz}{x^2+z^2}+\frac{x^2+z^2}{xz}\right)+\frac{3}{4}\)

\(\ge2.\frac{1}{2}+\frac{2.1}{2}+\frac{2.1}{2}+\frac{3}{4}=\frac{15}{4}\)(dpcm)

dau = xay ra khi x=y=z=1/3

Áp dụng BĐT Cô - si cho 3 số không âm:

\(1+x^3+y^3\ge3\sqrt[3]{1.x^3y^3}=3xy\Rightarrow\frac{\sqrt{1+x^3+y^3}}{xy}\ge\frac{\sqrt{3}}{\sqrt{xy}}\)

Tương tự ta có: \(\frac{\sqrt{1+y^3+z^3}}{yz}\ge\frac{\sqrt{3}}{\sqrt{yz}}\);\(\frac{\sqrt{1+z^3+x^3}}{zx}\ge\frac{\sqrt{3}}{\sqrt{zx}}\)

Cộng các vế của các BĐT trên, ta được:

\(\frac{\sqrt{1+x^3+y^3}}{xy}\)\(+\frac{\sqrt{1+y^3+z^3}}{yz}\)\(+\frac{\sqrt{1+z^3+x^3}}{zx}\ge\)\(\frac{\sqrt{3}}{\sqrt{xy}}\)\(+\frac{\sqrt{3}}{\sqrt{yz}}\)\(+\frac{\sqrt{3}}{\sqrt{zx}}\)

Tiếp tục áp dụng Cô - si:

\(BĐT\ge3\sqrt[3]{\frac{\sqrt{3}}{\sqrt{xy}}.\frac{\sqrt{3}}{\sqrt{yz}}.\frac{\sqrt{3}}{\sqrt{zx}}}=3\sqrt{3}\)

Vậy \(\frac{\sqrt{1+x^3+y^3}}{xy}\)\(+\frac{\sqrt{1+y^3+z^3}}{yz}\)\(+\frac{\sqrt{1+z^3+x^3}}{zx}\ge3\sqrt{3}\)

(Dấu "="\(\Leftrightarrow x=y=z=1\))

\(x^3+y^3+1=x^3+y^3+xyz\ge xy\left(x+y\right)+xyz=xy\left(x+y+z\right)\)

Tương tự:

\(y^3+z^3+1\ge yz\left(x+y+z\right);z^3+x^3+1\ge zx\left(x+y+z\right)\)

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

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

\(\ge\sqrt{3\sqrt[3]{xyz}}\cdot3\sqrt[3]{\frac{1}{\sqrt{xy}\cdot\sqrt{yz}\cdot\sqrt{zx}}}=3\sqrt{3}\)

Dấu "=" xảy ra tại \(x=y=z=1\)

\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge x;\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)

\(\Rightarrow VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{4}.2=\frac{3}{2}\)

Từ \(xy+yz+xz=xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)

Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\rightarrow\left(a,b,c\right)\) thì có

\(\frac{c^3}{\left(a+1\right)\left(b+1\right)}+\frac{b^3}{\left(a+1\right)\left(c+1\right)}+\frac{a^3}{\left(b+1\right)\left(c+1\right)}\ge\frac{1}{16}\)\(\forall\hept{\begin{cases}a+b+c=1\\a,b,c>0\end{cases}}\)

Áp dụng BĐT AM-GM ta có:

\(\frac{a^3}{\left(b+1\right)\left(c+1\right)}+\frac{b+1}{64}+\frac{c+1}{64}\ge\frac{3a}{16}\)

Tương tự cho 2 BĐT còn lại rồi cộng theo vế

\(VT+\frac{2\left(a+b+c+3\right)}{64}\ge\frac{3\left(a+b+c\right)}{16}\Leftrightarrow VT\ge\frac{1}{16}\)

Khi \(a=b=c=\frac{1}{3}\Leftrightarrow x=y=z=1\)

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

BaBie làm cái chi đây 

Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\Leftrightarrow xy+yz+zx=xyz\)

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

Bình phương vế trái : 

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

\(=\left(x+y+z+xy+yz+zx\right)+2\left(\sqrt{x+yz}.\sqrt{y+zx}+\sqrt{y+zx}.\sqrt{z+xy}+\sqrt{z+xy}.\sqrt{x+yz}\right)\)Bình phương vế phải : 

\(\left(\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2=\left(xyz+x+y+z\right)+2\left(x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)

Suy ra cần phải chứng minh : \(\sqrt{x+yz}.\sqrt{y+zx}+\sqrt{y+zx}.\sqrt{z+xy}+\sqrt{z+xy}.\sqrt{x+yz}\ge x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}+\sqrt{x}+\sqrt{y}+\sqrt{z}\)(*)

Thật vậy, theo bđt Bunhiacopxki ta có : \(\sqrt{x+yz}.\sqrt{y+zx}\ge\sqrt{xy}+z\sqrt{xy}\)

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

\(\sqrt{z+xy}.\sqrt{x+yz}\ge\sqrt{xz}+y\sqrt{xz}\)

Cộng các bđt trên theo vế ta chứng minh được (*) đúng.

Vậy bđt ban đầu được chứng minh.

Ý tưởng khác

Cũng từ giả thiết suy ra \(xyz=xy+yz+xz\)

Suy ra \(\sqrt{x+yz}=\sqrt{\frac{x^2+xyz}{x}}=\sqrt{\frac{x^2+xy+yz+xz}{x}}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{x}}\)

Theo BĐT Cauchy-Schwarz ta có \(\sqrt{\left(x+y\right)\left(x+z\right)}\ge x+\sqrt{yz}\) do đó:

\(\sqrt{x+yz}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{x}}\ge\frac{x+\sqrt{yz}}{x}=\sqrt{x}+\sqrt{\frac{yz}{x}}\)

Tương tự cho 2 BĐT còn lại \(\sqrt{y+xz}\ge\sqrt{y}+\sqrt{\frac{xz}{y}};\sqrt{z+xy}\ge\sqrt{z}+\sqrt{\frac{xy}{z}}\)

Cộng theo vế 3 BĐT được \(VT\ge\sqrt{x}+\sqrt{\frac{yz}{x}}+\sqrt{y}+\sqrt{\frac{xz}{y}}+\sqrt{z}+\sqrt{\frac{xy}{z}}\)

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

\(\Leftrightarrow VT\ge\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{xyz}\) (Đpcm)

bằng 3