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 \(x=y=z=1\)

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 \(\Leftrightarrow a=b=c=1\)

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 \(\Leftrightarrow x=y=z=1\)

các bạn ơi giúp mình với

\(\sqrt{4-x}\ge0\) với mọi x thuộc TXĐ nên \(y=\sqrt{4-x}+\sqrt{3}\ge\sqrt{3}\)

Đáp án D

Theo BĐT Cô - si: 

\(\sqrt{\frac{y+z}{x}.1}\le\left(\frac{y+z}{x}+1\right):2=\frac{x+y+z}{2x}\Rightarrow\sqrt{\frac{x}{y+z}}\ge\frac{2x}{x+y+z}\).  Bạn làm tương tự và cộng từng vế sau đó CM không xảy ra dấu bằng

2/

a/ \(\sqrt{a}+\frac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\frac{1}{\sqrt{a}}}=2\), dấu "=" khi \(a=1\)

b/ \(a+b+\frac{1}{2}=a+\frac{1}{4}+b+\frac{1}{4}\ge2\sqrt{a.\frac{1}{4}}+2\sqrt{b.\frac{1}{4}}=\sqrt{a}+\sqrt{b}\)

Dấu "=" khi \(a=b=\frac{1}{4}\)

c/ Có lẽ bạn viết đề nhầm, nếu đề đúng thế này thì mình ko biết làm

Còn đề như vậy: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\) thì làm như sau:

\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\) ; \(\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\); \(\frac{1}{x}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\)

Cộng vế với vế ta được:

\(\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\ge\frac{2}{\sqrt{xy}}+\frac{2}{\sqrt{yz}}+\frac{2}{\sqrt{xz}}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\)

Dấu "=" khi \(x=y=z\)

d/ \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\frac{\left(\sqrt{3}-2\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}\)

\(=\frac{7+4\sqrt{3}}{3-4}-\frac{7-4\sqrt{3}}{3-4}=-7-4\sqrt{3}+7-4\sqrt{3}=-8\sqrt{3}\)

e/ \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=\frac{\left(a-b\right)\left(a+b-\sqrt{ab}\right)}{\sqrt{ab}}\)

\(=\frac{a^2-b^2}{\sqrt{ab}}-\left(a-b\right)\) (bạn chép đề sai)

@Akai Haruma Cô giúp em với ạ!!!

? Cosi thôi câu 1 2 phần II

a) \(A=\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}\)

\(A=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)

\(A=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

b) \(B=\dfrac{x\sqrt{y}-y\sqrt{x}}{x-2\sqrt{xy}+y}\)

\(B=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)

\(B=\dfrac{\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\)

c) \(C=\dfrac{3\sqrt{a}-2a-1}{4a-4\sqrt{a}+1}\)

\(C=\dfrac{-\left(2a-3\sqrt{a}+1\right)}{\left(2\sqrt{a}\right)^2-2\sqrt{a}\cdot2\cdot1+1^2}\)

\(C=\dfrac{-\left(\sqrt{a}-1\right)\left(2\sqrt{a}-1\right)}{\left(2\sqrt{a}-1\right)^2}\)

\(C=\dfrac{-\sqrt{a}+1}{2\sqrt{a}-1}\)

d) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)

\(D=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}\)

\(D=\sqrt{a}+2-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)

\(D=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)\)

\(D=0\)

2.

Áp dụng bất đẳng thức Bunhiacopxki :

\(\left(1+9^2\right)\left(x^2+\frac{1}{x^2}\right)\ge\left(x+\frac{9}{x}\right)^2\)

\(\Leftrightarrow82\cdot\left(x^2+\frac{1}{x^2}\right)\ge\left(x+\frac{9}{x}\right)^2\)

\(\Leftrightarrow\sqrt{82}\cdot\sqrt{x^2+\frac{1}{x^2}}\ge x+\frac{9}{x}\)

Tương tự ta cũng có :

\(\sqrt{82}\cdot\sqrt{y^2+\frac{1}{y^2}}\ge y+\frac{9}{y}\)

\(\sqrt{82}\cdot\sqrt{z^2+\frac{1}{z^2}}\ge z+\frac{9}{z}\)

Cộng theo vế của các bất đẳng thức ta được :

\(\sqrt{82}\cdot\left(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\right)\ge x+y+z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}\)

\(\Leftrightarrow\sqrt{82}\cdot P\ge x+\frac{9}{x}+y+\frac{9}{y}+z+\frac{9}{z}\)(1)

Mặt khác áp dụng bất đẳng thức Cauchy ta có :

\(x+\frac{9}{x}+y+\frac{9}{y}+z+\frac{9}{z}=81x+\frac{9}{x}+81y+\frac{9}{y}+81z+\frac{9}{z}-80x-80y-80z\)

\(\ge2\sqrt{\frac{81x\cdot9}{x}}+2\sqrt{\frac{81y\cdot9}{y}}+2\sqrt{\frac{81z\cdot9}{z}}-80\left(x+y+z\right)\)

\(\ge2\sqrt{729}+2\sqrt{729}+2\sqrt{729}-80\cdot1\)

\(=82\) (2)

Từ (1) và (2) suy ra \(\sqrt{82}\cdot P\ge82\)

\(\Leftrightarrow P\ge\sqrt{82}\) ( đpcm )

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{1}{3}\)

1.

Áp dụng bất đẳng thức Cauchy :

\(\frac{a^2+1}{a}+\frac{b^2+1}{b}+\frac{c^2+1}{c}\)

\(=a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\)

\(=9a+\frac{1}{a}+9b+\frac{1}{b}+9c+\frac{1}{c}-8a-8b-8c\)

\(\ge2\sqrt{\frac{9a}{a}}+2\sqrt{\frac{9b}{b}}+2\sqrt{\frac{9c}{c}}-8\left(a+b+c\right)\)

\(\ge3\cdot2\sqrt{9}-8=10\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)