Với mọi số thực ta luôn có:

`(x-y)^2>=0`

`<=>x^2-2xy+y^2>=0`

`<=>x^2+y^2>=2xy`

`<=>(x+y)^2>=4xy`

`<=>(x+y)^2>=16`

`<=>x+y>=4(đpcm)`

\(\dfrac{1}{x+3}+\dfrac{1}{y+3}=\dfrac{x+3+y+3}{\left(x+3\right)\left(y+3\right)}\)

\(=\dfrac{x+y+6}{3x+3y+13}\)(vì \(xy=4\))

=> \(\dfrac{x+y+6}{3x+3y+13}\)≤\(\dfrac{2}{5}\)

<=> \(5\left(x+y+6\right)\)≤\(2\left(3x+3y+13\right)\)

<=>\(6x+6y+26-5x-5y-30\)≥\(0\)

<=> \(x+y-4\)≥\(0\)

Áp dụng BĐT AM-GM \(\dfrac{a+b}{2}\)≥\(\sqrt{ab}\)

Ta có \(\dfrac{x+y}{2}\)≥\(\sqrt{xy}\)

<=>\(x+y\) ≥ 2\(\sqrt{xy}\)

=>2\(\sqrt{xy}-4\)≥\(0\)

<=> \(4-4\)≥0

<=>0≥0 ( Luôn đúng )

Vậy \(\dfrac{1}{x+3}+\dfrac{1}{y+3}\)≤\(\dfrac{2}{5}\)

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

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

Ta có: \(xy\le\frac{\left(x+y\right)^2}{4}\)(bđt cosi)

=> \(\frac{\left(x+y\right)^2}{4}\ge4\) <=> \(\left(x+y\right)^2\ge16\) <=> \(x+y\ge4\)

CM bđt tương đương: \(\frac{1}{x+3}+\frac{1}{y+3}\le\frac{2}{5}\) 

<=> \(\frac{5\left(x+3\right)+5\left(y+3\right)}{\left(y+3\right)\left(y+3\right)}\le2\)

<=> \(2\left(xy+3x+3y+9\right)\ge5x+5y+30\)

<=> \(2.4+6\left(x+y\right)+18-5\left(x+y\right)-30\ge0\)

<=> \(x+y-4\ge0\) (vì x + y \(\ge\)4)

<=> \(4-4\ge0\) (Luôn đúng) 

=> ĐPCM

1) \(21x^2+21y^2+z^2\)

\(=18\left(x^2+y^2\right)+z^2+3\left(x^2+y^2\right)\)

\(\ge9\left(x+y\right)^2+z^2+3.2xy\)

\(\ge2.3\left(x+y\right).z+6xy\)

\(=6\left(xy+yz+zx\right)=6.13=78\)

Dấu "=" xảy ra <=> x = y ; 3(x+y) = z; xy + yz + zx= 13 <=> x = y = 1; z= 6

2) \(x+y+z=3xyz\)

<=> \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=3\)

Đặt: \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\)=> ab + bc + ca = 3

Ta cần chứng minh: \(3a^2+b^2+3c^2\ge6\)

Ta có: \(3a^2+b^2+3c^2=\left(a^2+c^2\right)+2\left(a^2+c^2\right)+b^2\)

\(\ge2ac+\left(a+c\right)^2+b^2\ge2ac+2\left(a+c\right).b=2\left(ac+ab+bc\right)=6\)

Vậy: \(\frac{3}{x^2}+\frac{1}{y^2}+\frac{3}{z^2}\ge6\)

Dấu "=" xảy ra <=> a = c = \(\sqrt{\frac{3}{5}}\); \(b=2\sqrt{\frac{3}{5}}\)

khi đó: \(x=z=\sqrt{\frac{5}{3}};y=\sqrt{\frac{5}{3}}\)

helloo

Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+z\right)\left(x+y\right)\)

Khi đó BĐT <=>

 \(\frac{1}{\left(x+y\right)\left(x+z\right)}+\frac{1}{\left(y+z\right)\left(x+z\right)}+\frac{1}{\left(x+y\right)\left(y+z\right)}\ge\frac{2}{3}\left(\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}+...\right)\)

<=> \(\frac{x+y+z}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\frac{x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}}{\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}}\right)^3\)

<=>\(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge\frac{1}{3}\left(x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}\right)^3\)

<=> \(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\right)^3\)(1)

Xét \(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge\frac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)

<=> \(9\left[xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\right]\ge8\left(xy\left(x+y\right)+xz\left(x+z\right)+yz\left(y+z\right)+3xyz\right)\)

<=> \(xy\left(y+x\right)+yz\left(y+z\right)+xz\left(x+z\right)\ge6xyz\)

<=> \(x\left(y-z\right)^2+z\left(x-y\right)^2+y\left(x-z\right)^2\ge0\)luôn đúng

Khi đó (1) <=> 

\(\left(x+y+z\right).\frac{2\sqrt{2}}{3}.\sqrt{x+y+z}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+....\right)^3\) 

<=> \(\sqrt{2\left(x+y+z\right)}\ge\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\)

Áp dụng buniacopxki cho vế phải ta có 

\(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\le\sqrt{\left(x+y+z\right)\left(3-xy-yz-xz\right)}\)

                                                                                                       \(=\sqrt{2\left(x+y+z\right)}\)

=> BĐT được CM

Dấu bằng xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)

\(VT\ge\frac{9}{\Sigma_{cyc}\sqrt{xy+x+y}}\ge\frac{9}{\sqrt{\left(1+1+1\right)\left(2x+2y+2z+xy+yz+zx\right)}}\ge\frac{9}{\sqrt{3\left[6+\frac{\left(x+y+z\right)^2}{3}\right]}}=\sqrt{3}\)

 Đoạn cuối của cô Nguyễn Linh Chi em có 1 cách biến đổi tương đương cũng khá ngắn gọn ạ

\(RHS\ge2\cdot\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\)

Theo đánh giá của cô Nguyễn Linh Chi thì \(xy+yz+zx\ge x+y+z\ge3\)

Ta cần chứng minh:\(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\ge\frac{1}{2}\)

Thật vậy,BĐT tương đương với:

\(2\left(x+y+z\right)^2\ge x^2+y^2+z^2-x-y-z+18\)

\(\Leftrightarrow\left(x+y+z\right)^2+x+y+z-12\ge0\)

\(\Leftrightarrow\left(x+y+z+4\right)\left(x+y+z-3\right)\ge0\) ( luôn đúng với \(x+y+z\ge3\) )

=> đpcm

Áp dụng: \(AB\le\frac{\left(A+B\right)^2}{4}\)với mọi A, B

Ta có:

\(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\le\frac{\left(x+2+x^2-2x+4\right)^2}{4}\)

=> \(\sqrt{x^3+8}\le\frac{x^2-x+6}{2}\)

=> \(\frac{x^2}{\sqrt{x^3+8}}\ge\frac{2x^2}{x^2-x+6}\)

Tương tự 

=> \(\frac{x^2}{\sqrt{x^3+8}}+\frac{y^2}{\sqrt{y^3+8}}+\frac{z^2}{\sqrt{z^3+8}}\)

\(\ge\frac{2x^2}{x^2-x+6}+\frac{2y^2}{y^2-y+6}+\frac{2z^2}{z^2-z+6}\)

\(=2\left(\frac{x^2}{x^2-x+6}+\frac{y^2}{y^2-y+6}+\frac{z^2}{z^2-z+6}\right)\)

\(\ge2\frac{\left(x+y+z\right)^2}{x^2-x+6+y^2-y+6+z^2-z+6}\)

\(=2\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\)(1)

Ta có: \(x+y+z\le xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\) với mọi x, y, z 

=> \(\left(x+y+z\right)^2-3\left(x+y+z\right)\ge0\)

=> \(\left(x+y+z\right)\left(x+y+z-3\right)\ge0\)

=> \(x+y+z\ge3\)với mọi x, y, z dương

Và \(x^2+y^2+z^2=\left(x+y+z\right)^2-2\left(xy+yz+zx\right)\le\left(x+y+z\right)^2-2\left(x+y+z\right)\)

Do đó: \(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\)

\(\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2-3\left(x+y+z\right)+18}\)

Đặt: x + y + z = t ( t\(\ge3\))

Xét hiệu: \(\frac{t^2}{t^2-3t+18}-\frac{1}{2}=\frac{t^2+3t-18}{t^2-3t+18}=\frac{\left(t-3\right)\left(t+6\right)}{\left(t-\frac{3}{2}\right)^2+\frac{63}{4}}\ge0\)với mọi t \(\ge3\)

Do đó: \(\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2-3\left(x+y+z\right)+18}\ge\frac{1}{2}\)(2)

Từ (1); (2) 

=> \(\frac{x^2}{\sqrt{x^3+8}}+\frac{y^2}{\sqrt{y^3+8}}+\frac{z^2}{\sqrt{z^3+8}}\ge2.\frac{1}{2}=1\)

Dấu "=" xảy ra <=> x= y = z = 1