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

Lời giải:
Áp dụng BĐT AM-GM ta có:

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

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

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

Cauchy-Schwarz:

\(3=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}\Rightarrow x+y+z\geq 3(2)\)

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

\(\Rightarrow \left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)\leq 3(3)\)

Từ \((1);(2);(3)\Rightarrow \text{VT}\geq 3-\frac{1}{2}.3=\frac{3}{2}\)

Mặt khác: \(\text{VP}=\frac{1}{2}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\frac{3}{2}\)

Do đó \(\text{VT}\geq \text{VP}\) (đpcm)

Dấu "=" xảy ra khi $x=y=z=1$

a) \(\text{ }x^4+y^4\ge x^3y+xy^3\)

\(\Leftrightarrow x^4+y^4-x^3y-xy^3\ge0\)

\(\Leftrightarrow x^3\left(x-y\right)-y^3\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)(ĐPCM) 

*NOTE: chứng minh đc vì (x-y)^2  >= 0 ;  x^2  +xy +y^2 > 0

mình cũng làm đến nơi rồi nhưng sợ x^2+xy+y^2 chưa chắc lớn hơn 0 thanks bạn nhé

ta có \(\left(x-y\right)^2\ge0\)

<=> \(x^2+y^2\ge2xy\)

<=>\(x^2+y^2+2xy\ge4xy\)

<=>\(\left(x+y\right)^2\ge4xy\)

<=>\(\frac{x+y}{xy}\ge\frac{4}{x+y}\)

<=>\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

Áp Dụng Cosi 3 số Ta phân tích B thành :

\(B=\frac{x^3}{1+y}+\frac{1+y}{4}+\frac{1}{2}+\frac{y^3}{1+x}+\frac{1+x}{4}+\frac{1}{2}-\frac{1+y}{4}-\frac{1+x}{4}-1\)

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

Ta có

\(\frac{x^3}{1+y}+\frac{1+y}{4}+\frac{1}{2}\ge3\sqrt[3]{\frac{x^3}{1+y}.\frac{1+y}{4}.\frac{1}{2}}=\frac{3x}{2}\)

\(\frac{y^3}{1+x}+\frac{1+x}{4}+\frac{1}{2}\ge3\sqrt[3]{\frac{y^3}{1+x}.\frac{1+x}{4}.\frac{1}{2}}=\frac{3y}{2}\)

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

\(\frac{3y}{2}+\frac{3x}{2}-\left(\frac{1+y}{4}+\frac{1+x}{4}\right)-1=\frac{3y+3x}{2}-\frac{1+y+1+x}{4}-1=\frac{6x+6y-1-y-1-x}{4}\)

\(=\frac{5y+5x-2}{4}-1\)

Ta có 

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

mà xy=1

\(\Rightarrow x+y\ge2\)

\(\Rightarrow5\left(x+y\right)\ge10\)

\(\Rightarrow5x+5y-2\ge8\)

\(\Rightarrow\frac{5x+5y-2}{4}\ge2\)

\(\Rightarrow\frac{5x+5y-2}{4}-1\ge1\)

Mà \(B\ge\frac{5x+5y-2}{4}-1\)

\(\Rightarrow B\ge\frac{5x+5y-2}{4}-1\ge1\Rightarrow B\ge1\left(dpcm\right)\)

Chúc bạn học tốt nha 

T I C K nha

(x^3)/(1+y)=(x^3)/(1+y)+(1+y)/4+1/2-(1+y)/4-1/2

Áp dụng bất đẳng thức Cosy cho 3 số:(x^3)/(1+y)  (1+y)/4 và 1/2 ta có

(x^3)/(1+y) +(1+y)/4 +1/2 \(\ge3\sqrt[3]{\left(\frac{x^3}{4\cdot2}\right)}=\frac{3}{2}\cdot x\)

CMTT ta có B>=3/2*(x+y)-(1+y+1+x)/4-1=3/2*(x+y)-(2+x+y)/4-1

ta có x+y>=\(2\sqrt{xy}\)=2

~>B>=3/2*2-1-1=1~> ĐPCM

còn 1 cách khác nữa đó

tiếp tục câu 2,vì máy bị lỗi nên phải tách ra:

Ta có:\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

\(=\left(x+y+z\right)\left(\left(x+y+z\right)^2-3\left(xy+xz+yz\right)\right).\)

Dó đó:\(x^3+y^3+z^3-3xyz+2010\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(\left(x+y+z\right)^2-3\left(xy+yz+xz\right)+2010\right)\)

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

TỪ \(\left(1\right),\left(2\right)\)suy ra \(P\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\frac{1}{x+y+z}.\)

Dấu \(=\)xảy ra khi \(x=y=z=\frac{\sqrt{2010}}{3}\)

2)Ta có:

\(x\left(x^2-yz+2010\right)=x\left(x^2+xy+xz+1340\right)>0\)

Tương tự ta có:\(y\left(y^2-xz+2010\right)>0,z\left(z^2-xy+2010\right)>0\)

Áp dụng svac-xơ ta có:

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

\(\ge\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2010\left(x+y+z\right)}.\)(1)