\(P=\frac{1}{\left(x+y\right)\left(\left(x+y\right)^2-3xy\right)}+\frac{1}{xy}=\frac{1}{1-3xy}+\frac{1}{xy}=\frac{1}{1-3xy}+\frac{3}{3xy}\ge\frac{\left(1+\sqrt{3}\right)^2}{1-3xy+3xy}=4+2\sqrt{3}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=\frac{3+\sqrt{6\sqrt{3}-9}}{6}\\y=\frac{3-\sqrt{6\sqrt{3}-9}}{6}\end{matrix}\right.\) và hoán vị

Bài 1:

Theo BĐT AM-GM có :$(x+y+1)(x^2+y^2)+\dfrac{4}{x+y}\geq (x+y+1).2xy+\dfrac{4}{x+y}=2(x+y+1)+\dfrac{4}{x+y}=(x+y)+(x+y)+\dfrac{4}{x+y}+2\geq 2\sqrt{xy}+2\sqrt{(x+y).\dfrac{4}{x+y}}+2=2+4+2=8$(đpcm)

Dấu \(=\) xảy ra khi \(x=y, xy=1\) và \(x+y=2\) hay \(x=y=1\)

Bài 1:

Áp dụng BĐT Cô-si cho các số dương:

\(x^2+y^2\geq 2xy=2\Rightarrow (x+y+1)(x^2+y^2)+\frac{4}{x+y}\geq 2(x+y+1)+\frac{4}{x+y}(1)\)

Tiếp tục áp dụng BĐT Cô-si:

\(2(x+y+1)+\frac{4}{x+y}=(x+y+2)+[(x+y)+\frac{4}{x+y}]\)

\(\geq (2\sqrt{xy}+2)+2\sqrt{(x+y).\frac{4}{x+y}}=(2+2)+4=8(2)\)

Từ \((1);(2)\Rightarrow (x+y+1)(x^2+y^2)+\frac{4}{x+y}\geq 8\) (đpcm)

Dấu "=" xảy ra khi $x=y=1$

Bài 2:

Vì $xyz=1$ nên:

\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}+\frac{3}{x+y+z}=\frac{z+x+y}{xyz}+\frac{3}{x+y+z}=x+y+z+\frac{3}{x+y+z}\)

Áp dụng BĐT Cô-si cho các số dương:
\(\frac{x+y+z}{3}+\frac{3}{x+y+z}\geq 2(1)\)

\(\frac{2}{3}(x+y+z)\geq \frac{2}{3}.3\sqrt[3]{xyz}=\frac{2}{3}.3=2(2)\)

Từ \((1);(2)\Rightarrow \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}+\frac{3}{x+y+z}\geq 2+2=4\)

Ta có đpcm.

Dấu "=" xảy ra khi $x=y=z=1$

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 \(\Leftrightarrow\)\(x=y=\frac{1}{2}\)

Ta có : \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\ge xy\left(x+y\right)\)

\(\Rightarrow1+x^3+y^3\ge xyz+xy\left(x+y\right)=xy\left(x+y+z\right)\ge3xy\sqrt[3]{xyz}=3xy\)

\(\frac{\sqrt{1+x^3+y^3}}{xy}\ge\frac{\sqrt{3xy}}{xy}=\sqrt{\frac{3}{xy}}\)

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

\(\Rightarrow A\ge\sqrt{3}\left(\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{xz}}\right)\ge3\sqrt{3}\sqrt{\frac{1}{\sqrt{x^2y^2z^2}}}=3\sqrt{3}\)

Ta có: \(x^3+y^3\ge xy\left(x+y\right)\Rightarrow1+x^3+y^3\ge xyz+xy\left(x+y\right)\)

\(=xy\left(x+y+z\right)\ge3xy\sqrt[3]{xyz}=3xy\)(vì xyz = 1)

\(\Rightarrow\frac{\sqrt{1+x^3+y^3}}{xy}=\frac{\sqrt{3xy}}{xy}=\sqrt{\frac{3}{xy}}\)

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

Cộng vế với vế, ta được:

\(BĐT=\sqrt{3}\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}\right)\)

\(\ge3\sqrt{3}\sqrt[3]{\frac{1}{\sqrt{x^2y^2z^2}}}=3\sqrt{3}\)

(Dấu "="\(\Leftrightarrow x=y=z=1\))

\(VT-VP=\Sigma_{cyc}\frac{\frac{1}{2}\left(x+y+1\right)\left(x-y\right)^2}{xy\left(\sqrt{x^3+y^3+1}+\sqrt{3xy}\right)}+\Sigma_{cyc}\frac{\left(x-1\right)^2}{xy\left(\sqrt{x^3+y^3+1}+\sqrt{3xy}\right)}\)

Tìm x : 

a) ( x - 15 ) . 35 = 0 

               x - 15 = 0 : 35

               x - 15 = 0  

                      x = 0 + 15

                      x = 15

b) 32 ( x - 10 ) = 32 

              x - 10 = 32 : 32

              x - 10 = 1

                     x = 1 + 10

                     x = 11

\(\frac{1}{x^3+y^3}+\frac{1}{xy}=\frac{1}{x^2-xy+y^2}+\frac{1}{x}+\frac{1}{y}=1+\frac{3xy}{x^3+y^3}+1+\frac{x}{y}+1+\frac{y}{x}\ge5\)

Ta có: \(x^3+y^3=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)

Áp dụng BĐT CAuchy-Schwarz ta có:

\(VT=\frac{1}{x^3+y^3}+\frac{1}{xy}=\frac{x+y}{\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]}+\frac{1}{xy}\)

\(=\frac{1}{1-3xy}+\frac{3}{3xy}=\frac{1}{1-3xy}+\frac{\sqrt{3}^2}{3xy}\)

\(\ge\frac{\left(1+\sqrt{3}\right)^2}{1-3xy+3xy}=\left(1+\sqrt{3}\right)^2=4+2\sqrt{3}\)

\(P=\frac{1}{x^3+y^3}+\frac{1}{xy}\)

Ta có:

\(x+y=1\Rightarrow\left(x+y\right)^3=1\)

\(\Rightarrow x^3+y^3+3xy\left(x+y\right)=1\)

\(\Rightarrow x^3+y^3+3xy=1\)

\(\Rightarrow P=\frac{x^3+y^3+3xy}{x^3+y^3}+\frac{x^3+y^3+3xy}{xy}\)\(=4+\frac{3xy}{x^3+y^3}+\frac{x^3+y^3}{xy}\left(1\right)\)

Áp dụng Bđt Cô si ta có:

\(\frac{3xy}{x^3+y^3}+\frac{x^3+y^3}{xy}\ge2\sqrt{\frac{3xy}{x^3+y^3}\cdot\frac{x^3+y^3}{xy}}=2\sqrt{3}\)

\(\Rightarrow P\ge4+2\sqrt{3}\)(Đpcm)

Dấu = khi \(\hept{\begin{cases}x+y=1\\x^3+y^3=\sqrt{3xy}\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=1\\1-3xy=\sqrt{3xy}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+y=1\\3\sqrt{xy}=\frac{-1+\sqrt{5}}{2}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x+y=1\\xy=\frac{6-2\sqrt{5}}{12}\end{cases}}\)

\(\Leftrightarrow x^2-x+\frac{6-2\sqrt{5}}{12}=0\)\(\Leftrightarrow x,y=\frac{1\pm\sqrt{\frac{2\sqrt{5}-3}{3}}}{2}\)

chiu

tk nhe

xin do

bye