với 2 số dương a,b ta luôn có

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\):\(\left(a+b\right)^2\ge4ab\)

Áp dụng vào bài toán, ta có

\(\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{2}{2xy}\)

\(=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{2}{4xy}\ge\frac{4}{\left(x+y\right)^2}+\frac{2}{\left(x+y\right)^2}=6\)(vì x+y=1)

Ta có: x²+y²≤(x+y)²/2 => 1/(x²+y²)≥2/(x+y)²=2

xy≤(x+y)²/4 => 1/xy≥4/(x+y)²=4

=>1/(x²+y²)+1/xy≥2+4=6

Dấu "=" xảy ra khi x=y=1/2

Xét: \(\frac{1-x^2}{x+yz}+\frac{1-y^2}{y+xz}+\frac{1-z^2}{z+xy}\)

Thay thế \(x+y+z=1\)

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

Áp dụng hằng đẳng thức hiệu 2 bình phương: \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)

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

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

Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm

\(\Rightarrow\left\{\begin{matrix}\left(x+y\right)\left(x+z\right)\le\left(\frac{2x+y+z}{2}\right)^2=\frac{\left(2x+y+z\right)^2}{4}\\\left(x+y\right)\left(y+z\right)\le\left(\frac{x+2y+z}{2}\right)^2=\frac{\left(x+2y+z\right)^2}{4}\\\left(x+z\right)\left(y+z\right)\le\left(\frac{x+y+2z}{2}\right)^2=\frac{\left(x+y+2z\right)^2}{4}\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}\frac{\left(y+z\right)\left(2x+y+z\right)}{\left(x+y\right)\left(x+z\right)}\ge\frac{4\left(y+z\right)\left(2x+y+z\right)}{\left(2x+y+z\right)^2}=\frac{4\left(y+z\right)}{2x+y+z}\\\frac{\left(x+z\right)\left(x+2y+z\right)}{\left(x+y\right)\left(y+z\right)}\ge\frac{4\left(x+z\right)\left(x+2y+z\right)}{\left(x+2y+z\right)^2}=\frac{4\left(x+z\right)}{x+2y+z}\\\frac{\left(x+y\right)\left(x+y+2z\right)}{\left(x+z\right)\left(y+z\right)}\ge\frac{4\left(x+y\right)\left(x+y+2z\right)}{\left(x+y+2z\right)^2}=\frac{4\left(x+y\right)}{x+y+2z}\end{matrix}\right.\)

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

\(\Rightarrow VT\ge4\left(\frac{y+z}{2x+y+z}+\frac{x+z}{x+2y+z}+\frac{x+y}{x+y+2z}\right)\)

Ta có: \(x+y+z=1\)

\(\Rightarrow\left\{\begin{matrix}y+z=1-x\\x+z=1-y\\x+y=1-z\end{matrix}\right.\) ( 1 )

\(\Rightarrow\left\{\begin{matrix}2x+y+z=1+x\\x+2y+z=1+y\\x+y+2z=1+z\end{matrix}\right.\) ( 2 )

Từ ( 1 ) và ( 2 )

\(\Rightarrow VT\ge4\left(\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z}\right)\)

\(\Rightarrow VT\ge4\left(\frac{1+x-2x}{1+x}+\frac{1+y-2y}{1+y}+\frac{1+z-2z}{1+z}\right)\)

\(\Rightarrow VT\ge4\left[3-\left(\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\right)\right]\)

\(\Rightarrow VT\ge12-4\left(\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\right)\)

Chứng minh rằng \(12-4\left(\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\right)\ge6\)

\(\Leftrightarrow4\left(\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\right)\le6\)

\(\Leftrightarrow\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\le\frac{3}{2}\)

\(\Leftrightarrow\frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z}\le\frac{3}{4}\)

\(\Leftrightarrow\frac{1+x-1}{1+x}+\frac{1+y-1}{1+y}+\frac{1+z-1}{1+z}\le\frac{3}{4}\)

\(\Leftrightarrow1-\frac{1}{1+x}+1-\frac{1}{1+y}+1-\frac{1}{1+z}\le\frac{3}{4}\)

\(\Leftrightarrow3-\left(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\right)\le\frac{3}{4}\)

Áp dụng bất đẳng thức cộng mẫu số

\(\Rightarrow\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge\frac{\left(1+1+1\right)^2}{3+x+y+z}=\frac{9}{4}\)

\(\Rightarrow3-\left(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\right)\le3-\frac{9}{4}\)

\(\Rightarrow3-\left(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\right)\le\frac{3}{4}\) ( đpcm )

Vì \(12-4\left(\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\right)\ge6\)

\(\Rightarrow VT\ge6\)

\(\Leftrightarrow\)\(\frac{1-x^2}{x+yz}+\frac{1-y^2}{y+xz}+\frac{1-z^2}{z+xy}\ge6\) ( đpcm )

Cách khác:

\(A=\frac{1-x^2}{x+yz}+\frac{1-y^2}{y+xz}+\frac{1-z^2}{z+xy}=\frac{1-x^2}{x(x+y+z)+yz}+\frac{1-y^2}{y(x+y+z)+xz}+\frac{1-z^2}{z(x+y+z)+xy}\)

\(\Leftrightarrow A=\frac{1-x^2}{(x+y)(x+z)}+\frac{1-y^2}{(y+z)(y+x)}+\frac{1-z^2}{(z+x)(z+y)}=\frac{2(x+y+z)-[xy(x+y)+yz(y+z)+xz(x+z)]}{(x+y)(y+z)(x+z)}\)

Có \(A\geq 6\Leftrightarrow 2-[xy(x+y)+yz(y+z)+xz(x+z)]\ge 6(x+y)(y+z)(x+z)\)

\(\Leftrightarrow 2+9xyz\geq 7(x+y+z)(xy+yz+xz)\)

\(\Leftrightarrow 2+9xyz\geq 7(xy+yz+xz)\) \((\star)\)

Theo BĐT Schur bậc 3 kết hợp AM-GM:

\(xyz\geq (x+y-z)(y+z-x)(x+z-y)=(1-2x)(1-2y)(1-2z)\)

\(\Leftrightarrow 9xyz\geq 4(xy+yz+xz)-1\)

\(\Rightarrow 2+9(xy+yz+xz)\geq 1+4(xy+yz+xz)=(x+y+z)^2+4(xy+yz+xz)\)\(\geq 7(xy+yz+xz)\)

Do đó \((\star)\) được CM. Bài toán hoàn tất. Dấu bằng xảy ra khi \(x=y=z=\frac{1}{3}\)

Áp dung BĐT \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\left(a,b,c>0\right)\)

\(=>x,y,z>0\left(taco\right)\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\ge\frac{9}{xy+yz+xz}\)

\(=>P\ge\frac{1}{x^2+y^2+z^2}+\frac{9}{xy+yz+xz}\)

\(=>P\ge\left(\frac{1}{x^2+y^2+z^2}+\frac{1}{xy+yz+zx}+\frac{1}{xy+yz+zx}\right)+\frac{7}{xy+yz+xz}\)

\(\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2zx}+\frac{7}{xy+yz+zx}\)

\(=\frac{9}{\left(x+y+z\right)^2}+\frac{7}{xy+yz+xz}\ge\frac{9}{\left(x+y+z\right)^2}+\frac{21}{\left(x+y+z\right)^2}\ge30\)

do \(3\left(xy+yz+zx\right)\le\left(x+y+z\right)^2and\left(x+y+z=1\right)\)

dấu = xảy ra khi x=y=z=1/3

zậy...........

\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\)

\(=\frac{x}{z}+\frac{y}{z}+\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}\)

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

Áp dụng BĐT AM-GM ta có:

\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\ge2.\sqrt{\frac{x}{z}.\frac{z}{x}}+2.\sqrt{\frac{x}{y}.\frac{y}{x}}+2.\sqrt{\frac{y}{z}.\frac{z}{y}}=2+2+2=6\)

                                                                                                                           đpcm

Svac-xơ 

\(VT=\left(\frac{x+y}{z}+1\right)+\left(\frac{y+z}{x}+1\right)+\left(\frac{z+x}{y}+1\right)-3\)

\(VT=\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{z}-3=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3\)

\(\ge\left(x+y+z\right).\frac{\left(1+1+1\right)^2}{x+y+z}-3=9-3=6\)

Các bạn chú ý khi chứng minh một bất đẳng thức có dấu bằng thì các bạn phải chỉ ra dấu "=" xảy ra khi nào?

(Dấu "=" xảy ra khi và chỉ khi x=y=z)

Đề sai rồi phải là \(\frac{x+y}{2}\ge\sqrt{xy}\) chứ!

Ta có: \(\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\forall x;y\)

         \(\Leftrightarrow x-2\sqrt{xy}+y\ge0\)

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

        \(\Leftrightarrow\frac{x+y}{2}\ge\sqrt{xy}\)

Dấu "=" xảy ra khi x = y

Vậy .....

Tk nha!

mình chỉnh lại đề: CMR:  \(\frac{x+y}{2}\ge\sqrt{xy}\)     \(\forall x;y>0\)

           Bài làm

Áp dụng BĐT AM-GM ta có:

       \(\frac{x+y}{2}\ge\frac{2\sqrt{xy}}{2}=\sqrt{xy}\)

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

\(x^2+y^2\ge2xy\)

\(\Rightarrow x^2+y^2+2xy\ge4xy\)

\(\Rightarrow\left(x+y\right)^2\ge4xy\)

\(\Rightarrow x+y\ge\sqrt{4xy}\)

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

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

Xin lỗi bạn, mình mới học lớp 6 nén không bít trả lời cau hỏi này.

Xin lỗi nha!

Áp dụng bất đẳng thức \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)  được

\(VT\ge\frac{4}{\left(x+y\right)^2}\ge4\)

Dấu "=" xảy ra khi x = y = 1/2

Vậy ...........

Cũng ko hẳn là cách khác nhưng xem cho vui v :) 

\(\frac{1}{x^2+xy}+\frac{1}{y^2+xy}=\frac{1}{x\left(x+y\right)}+\frac{1}{y\left(x+y\right)}\ge\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\ge4\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=\frac{1}{2}\)