AD=\(\sqrt{3}\)R

AE=(1/4)AD=\(\frac{\sqrt{3}}{4}\)R

DE=\(\frac{3\sqrt{3}}{4}\)R

đk: \(\hept{\begin{cases}x\ne0\\x^2\ge\sqrt{\frac{5}{6}}\end{cases}}\)

Ta có: \(x^2\ge\sqrt{\frac{5}{6}}\Rightarrow\hept{\begin{cases}\frac{5}{x^2}>0\\6x^2-1>0\end{cases}}\)

Áp dụng BĐT Cauchy ta có: \(\sqrt{30-\frac{5}{x^2}}=\sqrt{\frac{5}{x^2}\left(6x^2-1\right)}\le\frac{\frac{5}{x^2}+6x^2-1}{2}\) (1)

Mà \(\sqrt{6x^2-\frac{5}{x^2}}=\sqrt{\left(6x^2-\frac{5}{x^2}\right)\cdot1}\le\frac{6x^2-\frac{5}{x^2}+1}{2}\) (2)

Cộng vế (1) và (2) lại ta được:

\(\sqrt{30-\frac{5}{x^2}}+\sqrt{6x^2-\frac{5}{x^2}}\le\frac{12x^2}{2}=6x^2\) 

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\frac{5}{x^2}=6x^2-1\\6x^2-\frac{5}{x^2}=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Vậy \(x\in\left\{-1;1\right\}\)

Theo giả thiết, ta có: \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=2019\Rightarrow x+y+z=2019xyz\)

Xét \(\sqrt{2019x^2+1}=\sqrt{\frac{x^2+xy+xz}{yz}+1}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{yz}}\le\frac{\frac{x+y}{y}+\frac{x+z}{z}}{2}=\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)+1\)\(\Rightarrow\frac{x^2+1+\sqrt{2019x^2+1}}{x}\le\frac{x^2+1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)+1}{x}=x+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)+\frac{2}{x}\)

Tương tự, ta có: \(\frac{y^2+1+\sqrt{2019y^2+1}}{y}\le y+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)+\frac{2}{y}\); \(\frac{z^2+1+\sqrt{2019z^2+1}}{z}\le z+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{2}{z}\)

Cộng theo vế ba bất đẳng thức trên, ta được: \(\frac{x^2+1+\sqrt{2019x^2+1}}{x}+\frac{y^2+1+\sqrt{2019y^2+1}}{y}+\frac{z^2+1+\sqrt{2019z^2+1}}{z}\le\left(x+y+z\right)+3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=2019xyz+3.\frac{xy+yz+zx}{xyz}\le2019xyz+3.\frac{\frac{\left(x+y+z\right)^2}{3}}{xyz}=2019xyz+\frac{\left(x+y+z\right)^2}{xyz}=2019xyz+2019^2xyz=2019.2020xyz\)Đẳng thức xảy ra khi \(x=y=z=\frac{1}{\sqrt{673}}\)

a) \(ĐK:y-2x+1\ge0;4x+y+5\ge0;x+2y-2\ge0,x\le1\)

Th1: \(\hept{\begin{cases}y-2x+1=0\\3-3x=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}0=0\\-1=\sqrt{10}-1\end{cases}}\)(không thỏa mãn)

Th2: \(x,y\ne1\)

\(2x^2-y^2+xy-5x+y+2=\sqrt{y-2x+1}-\sqrt{3-3x}\)\(\Leftrightarrow\left(x+y-2\right)\left(2x-y-1\right)=\frac{x+y-2}{\sqrt{y-2x+1}+\sqrt{3-3x}}\)\(\Leftrightarrow\left(x+y-2\right)\left(\frac{1}{\sqrt{y-2x+1}+\sqrt{3-3x}}+y-2x+1\right)=0\)

Dễ thấy \(\frac{1}{\sqrt{y-2x+1}+\sqrt{3-3x}}+y-2x+1>0\)nên x + y - 2 = 0

Thay y = 2 - x vào phương trình \(x^2-y-1=\sqrt{4x+y+5}-\sqrt{x+2y-2}\), ta được: \(x^2+x-3=\sqrt{3x+7}-\sqrt{2-x}\)\(\Leftrightarrow x^2+x-2=\sqrt{3x+7}-1+2-\sqrt{2-x}\)\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=\frac{3\left(x+2\right)}{\sqrt{3x+7}+1}+\frac{x+2}{2+\sqrt{2-x}}\)\(\Leftrightarrow\left(x+2\right)\left(\frac{3}{\sqrt{3x+7}+1}+\frac{1}{2+\sqrt{2-x}}+1-x\right)=0\)

Vì \(x\le1\)nên\(\frac{3}{\sqrt{3x+7}+1}+\frac{1}{2+\sqrt{2-x}}+1-x>0\)suy ra x = -2 nên y = 4

Vậy nghiệm của hệ phương trình là (x;y) = (-2;4)

b) \(\hept{\begin{cases}x^2+y^2=5\\x^3+2y^3=10x-10y\end{cases}}\Leftrightarrow\hept{\begin{cases}2\left(x^2+y^2\right)=10\left(1\right)\\x^3+2y^3=10\left(x-y\right)\left(2\right)\end{cases}}\)

Thay (1) vào (2), ta được: \(x^3+2y^3=2\left(x^2+y^2\right)\left(x-y\right)\Leftrightarrow\left(2y-x\right)\left(x^2+2y^2\right)=0\)

* Th1: \(x^2+2y^2=0\)(*)

Mà \(x^2\ge0\forall x;2y^2\ge0\forall y\Rightarrow x^2+2y^2\ge0\)nên (*) xảy ra khi x = y = 0 nhưng cặp nghiệm này không thỏa mãn hệ

* Th2: 2y - x = 0 suy ra x = 2y thay vào (1), ta được: \(y^2=1\Rightarrow y=\pm1\Rightarrow x=\pm2\) 

Vậy hệ có 2 nghiệm \(\left(x,y\right)\in\left\{\left(2;1\right);\left(-2;-1\right)\right\}\)

2. \(BĐT\Leftrightarrow\frac{1}{1+\frac{2}{a}}+\frac{1}{1+\frac{2}{b}}+\frac{1}{1+\frac{2}{c}}\ge1\)

Đặt\(\frac{2}{a}=x;\frac{2}{b}=y;\frac{2}{c}=z\)thì \(\hept{\begin{cases}x,y,z>0\\xyz=8\end{cases}}\)

Ta cần chứng minh \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge1\Leftrightarrow\left(yz+y+z+1\right)+\left(zx+z+x+1\right)+\left(xy+x+y+1\right)\ge xyz+\left(xy+yz+zx\right)+\left(x+y+z\right)+1\)\(\Leftrightarrow x+y+z\ge6\)(Đúng vì \(x+y+z\ge3\sqrt[3]{xyz}=6\))

Đẳng thức xảy ra khi x = y = z = 2 hay a = b = c = 1

3. Ta có: \(a+b+c\le\sqrt{3}\Rightarrow\left(a+b+c\right)^2\le3\)

Ta có đánh giá quen thuộc \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

Từ đó suy ra \(ab+bc+ca\le1\)

\(A=\frac{\sqrt{a^2+1}}{b+c}+\frac{\sqrt{b^2+1}}{c+a}+\frac{\sqrt{c^2+1}}{a+b}\ge\frac{\sqrt{a^2+ab+bc+ca}}{b+c}+\frac{\sqrt{b^2+ab+bc+ca}}{c+a}+\frac{\sqrt{c^2+ab+bc+ca}}{a+b}\)\(=\frac{\sqrt{\left(a+b\right)\left(a+c\right)}}{b+c}+\frac{\sqrt{\left(b+a\right)\left(b+c\right)}}{c+a}+\frac{\sqrt{\left(c+a\right)\left(c+b\right)}}{a+b}\ge3\sqrt[3]{\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=3\)Đẳng thức xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)