Câu hỏi của Đặng Kim Anh

Question

1 ) Cho a,b,c >0 và abc= 1.CMR:

$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3$

2 ) Cho x,y,z > 0 và x+y+z=3

CMR : \(\frac{x}{x+\sqrt{3x+yz}}+\fr...

Nguyễn Việt Lâm · Accepted Answer

a/ $VT\ge\frac{\left(\sqrt{b}+\sqrt{c}\right)^2}{2\sqrt{a}}+\frac{\left(\sqrt{c}+\sqrt{a}\right)^2}{2\sqrt{b}}+\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\sqrt{c}}$

$VT\ge\frac{\left(\sqrt{b}+\sqrt{c}+\sqrt{c}+\sqrt{a}+\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}=2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)$

$VT\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b}+\sqrt{c}$

$VT\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[3]{\sqrt{abc}}=\sqrt{a}+\sqrt{b}+\sqrt{c}+3$

Dấu "=" xảy ra khi $a=b=c=1$

b/ $VT=\sum\frac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}=\sum\frac{x}{x+\sqrt{\left(x+y\right)\left(z+x\right)}}$

$VT\le\sum\frac{x}{x+\sqrt{xz}+\sqrt{xy}}=\sum\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1$

Dấu "=" xảy ra khi $x=y=z=1$Câu trả lời: a/ $VT\ge\frac{\left(\sqrt{b}+\sqrt{c}\right)^2}{2\sqrt{a}}+\frac{\left(\sqrt{c}+\sqrt{a}\right)^2}{2\sqrt{b}}+\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\sqrt{c}}$

$VT\ge\frac{\left(\sqrt{b}+\sqrt{c}+\sqrt{c}+\sqrt{a}+\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}=2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)$

$VT\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b}+\sqrt{c}$

$VT\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[3]{\sqrt{abc}}=\sqrt{a}+\sqrt{b}+\sqrt{c}+3$

Dấu "=" xảy ra khi $a=b=c=1$

b/ $VT=\sum\frac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}=\sum\frac{x}{x+\sqrt{\left(x+y\right)\left(z+x\right)}}$

$VT\le\sum\frac{x}{x+\sqrt{xz}+\sqrt{xy}}=\sum\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1$

Dấu "=" xảy ra khi $x=y=z=1$

Hoàng Thị Ánh Phương · Answer

Bài 1 :

Áp dụng BĐT Cô - si cho 2 số không âm ta có :

$VT=\Sigma_{cyc}\frac{b+c}{\sqrt{a}}\ge2\left(\Sigma_{cyc}\sqrt{\frac{bc}{a}}\right)$

$\Leftrightarrow\Sigma_{cyc}\frac{b+c}{\sqrt{a}}\ge\left(\sqrt{\frac{ca}{b}}+\sqrt{\frac{ab}{c}}\right)+\left(\sqrt{\frac{ab}{c}}+\sqrt{\frac{bc}{a}}\right)+\left(\sqrt{\frac{bc}{a}}+\sqrt{\frac{ca}{b}}\right)$

$\Leftrightarrow\Sigma_{cyc}\frac{b+c}{\sqrt{a}}\ge2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\ge\sqrt{a}+\sqrt{b}+\sqrt{c}$

$+3\sqrt[6]{abc}=\sqrt{a}+\sqrt{b}+\sqrt{c}+3$

Dấu " = " xảy ra $\Leftrightarrow a=b=c=1$Câu trả lời: Bài 1 :

Áp dụng BĐT Cô - si cho 2 số không âm ta có :

$VT=\Sigma_{cyc}\frac{b+c}{\sqrt{a}}\ge2\left(\Sigma_{cyc}\sqrt{\frac{bc}{a}}\right)$

$\Leftrightarrow\Sigma_{cyc}\frac{b+c}{\sqrt{a}}\ge\left(\sqrt{\frac{ca}{b}}+\sqrt{\frac{ab}{c}}\right)+\left(\sqrt{\frac{ab}{c}}+\sqrt{\frac{bc}{a}}\right)+\left(\sqrt{\frac{bc}{a}}+\sqrt{\frac{ca}{b}}\right)$

$\Leftrightarrow\Sigma_{cyc}\frac{b+c}{\sqrt{a}}\ge2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\ge\sqrt{a}+\sqrt{b}+\sqrt{c}$

$+3\sqrt[6]{abc}=\sqrt{a}+\sqrt{b}+\sqrt{c}+3$

Dấu " = " xảy ra $\Leftrightarrow a=b=c=1$

Hoàng Thị Ánh Phương · Answer

Bài 2 :

Ta có :

$\left(x-\sqrt{yz}\right)^2\ge0\Rightarrow x^2+yz\ge2x\sqrt{yz}$

( Dấu " = " $\Leftrightarrow x^2=yz$ )

Theo đề bài ta có : $x+y+z=3\Rightarrow3x+yz=\left(x+y+z\right)x+yz=x^2+yz+x\left(y+z\right)$

$\ge x\left(y+z\right)+2x\sqrt{yz}$

Suy ra $\sqrt{3x+yz}\ge\sqrt{x\left(y+z\right)+2x\sqrt{yz}}=\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)$

và $x+\sqrt{3x+yz}\ge\sqrt{x}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)$

$\Rightarrow\frac{x}{x+\sqrt{3x+yz}}\le\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}$

Tương tự ta cũng có : $\frac{y}{y+\sqrt{3y+zx}}\le\frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}};\frac{z}{z+\sqrt{3z+xy}}\le\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}$

Cộng từng vế của các BĐT trên , ta được :

$\frac{x}{x+\sqrt{3x+yz}}+\frac{y}{y+\sqrt{3y+zx}}+\frac{z}{z+\sqrt{3z+xy}}\le1$

Dấu " = " xảy ra $\Leftrightarrow x=y=z=1$Câu trả lời: Bài 2 :