nguyentinh

 Ta có ${\sqrt{\frac{a}{b+c}}}=\frac{\sqrt{a}}{\sqrt{b+c}}=\frac{a}{\sqrt{a\left ( b+c \right )}}\geq \frac{a}{\frac{1}{2}\left ( a+b+c \right )}=\frac{2a}{a+b+c}$(BĐT Cauchy)

Tương tự $\sqrt{\frac{b}{c+a}}\geq \frac{2b}{a+b+c};\sqrt{\frac{c}{a+b}}\geq \frac{2c}{a+b+c}$

$\Rightarrow \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\geq \frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c}=2$

Dấu = xảy ra $\Leftrightarrow \left\{\begin{matrix} a=b+c & & \\ b=c+a & & \\ c=a+b & & \end{matrix}\right.\Leftrightarrow a=b=c=0$ (Loại do $a,b,c>0$)

Vậy $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2$

Cho là $x^{2}+y^{2}+z^{2}=3$ thì ta có $P=\frac{x}{^{\sqrt[3]{yz}}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\geq3$.

Thật vậy

Áp dụng BĐT Cauchy: 

$P=\frac{x}{^{\sqrt[3]{yz}}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\geq 3\left ( \frac{x}{y+z+1}+\frac{y}{x+z+1}+\frac{z}{x+y+1}\right )$

$\Rightarrow \frac{P}{3}\geq \frac{x^{2}}{xy+xz+x}+\frac{y^{2}}{yz+yx+y}+\frac{z^{2}}{zx+zy+z}\geq \frac{\left ( x+y+z \right )^{2}}{2\left ( xy+yz+xz \right )+x+y+z}$( Cauchy- Schwarz)

Do $3\left ( x^{2}+y^{2}+z^{2} \right )\geq \left ( x+y+z \right )^{2}$

$\Rightarrow \left ( x+y+z \right )^{2}\leq \left ( x^{2}+y^{2}+z^{2} \right )^{2}$

$\Rightarrow x+y+z\leq x^{2}+y^{2}+z^{2}$

$\Rightarrow 2\left ( xy+yz+xz \right )+x+y+z\leq 2\left ( xy+yz+xz \right )+x^{2}+y^{2}+z^{2}=\left ( x+y+z \right )^{2}$

$\Rightarrow \frac{P}{3}\geq \frac{\left ( x+y+z \right )^{2}}{\left ( x+y+z \right )^{2}}=1$

$\Rightarrow P\geq 3$

Dấu = xảy ra $\Leftrightarrow x=y=z=1$

Bài 2:

Ta có $3\left ( \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \right )\geq \left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^{2}$

$\Rightarrow 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 2.3\left ( \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right )\geq 2\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^{2}$

$\Rightarrow 2\left ( \frac{1}{a}+ \frac{1}{b} +\frac{1}{c}\right )^{2}-\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )-1\leq 0$

$\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq 1$

Mà $\frac{1}{10a+b+c}=\frac{1}{a+a+a+a+a+a+a+a+a+a+b+c}\leq \frac{1}{144}.\left ( \frac{1}{a}+...+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )= \frac{1}{144}.\left ( \frac{10}{a}+\frac{1}{b}+\frac{1}{c} \right )$

Cmtt, Suy ra $\frac{1}{10a+b+c}+\frac{1}{a+10b+c}+\frac{1}{a+b+10c}\leq \frac{1}{144}.12.\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\leq \frac{1}{12}$

Dấu = xảy ra $\Leftrightarrow a=b=c=3$

$P=x^{2}+y^{2}+xy-5x-4y+2002$

$\Rightarrow 4P=4x^{2}+4y^{2}+4xy-20x-16y+8008$

$=4x^{2}+\left ( 4xy-20x \right )+4y^{2}-16y+8008$

$=4x^{2}+4x\left ( y-5 \right )+\left ( y^{2}-10y+25 \right )+3y^{2}-6y+3+7980$

$=\left ( 2x \right )^{2}+2.2x.\left ( y-5 \right )+\left ( y-5 \right )^{2}+3\left ( y-1 \right )^{2}+7980$

$=\left ( 2x+y-5 \right )^{2}+3\left ( y-1 \right )^{2}+7980\geq 7980 \forall x,y\in \mathbb{R}$

$\Rightarrow 4P\geq 7980\Rightarrow P\geq 1995$

Dấu '=' xảy ra $\Leftrightarrow \left\{\begin{matrix} & 2x+y-5=0\\ & y-1=0\ \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} & x=2\\ & y=1 \end{matrix}\right.$

Vậy $minP=1995$ tại $\left\{\begin{matrix} & x=2\\ & y=1 \end{matrix}\right.$

Câu 10 bạn sai rồi

$2\left ( x^{2}+y^{2} \right )\geq \left ( x+y \right )^{2}=16\Rightarrow x^{2}+y^{2}\geq 8$

$maxA=8 \Leftrightarrow x=y=2$

pt vô nghiệm 

$\Leftrightarrow \Delta '<0$

$\Leftrightarrow 4m^{4}-4m+\frac{13}{4}<0$

$\Leftrightarrow \left ( 4m^{4}-4m^{2}+1 \right )+\left ( 4m^{2}-4m+1\right)+\frac{5}{4}<0$

$\Leftrightarrow \left ( 2m^{2}-1 \right )^{2}+\left ( 2m-1 \right )^{2}+\frac{5}{4}<0$

(Loại)

Vậy không tồn tại giá trị của $m$ để phương trình đã cho vô nghiệm.