a, Đặt \(\sqrt[4]{a}=x;\sqrt[4]{b}=y.\)Bất đẳng thức ban đầu trở thành: \(\frac{2x^2y^2}{x^2+y^2}\le xy.\)

ta có : \(x^2+y^2\ge2xy\Rightarrow\frac{2x^2y^2}{x^2+y^2}\le\frac{2x^2y^2}{2xy}=xy.\)(đpcm ) 

dấu " = " xẩy ra khi x = y > 0 

vậy bất đăng thức ban đầu đúng. dấu " = " xẩy ra khi a = b >0

a) Sai đề.

\(\dfrac{a+b}{b^2}\sqrt[]{\dfrac{a^2b^4}{a^2+2ab+b^2}}=\dfrac{a+b}{b^2}.\dfrac{b^2\left|a\right|}{\left|a+b\right|}=\left|a\right|\)

b) Sai đề.

\(\dfrac{a\sqrt[]{b}+b\sqrt[]{a}}{\sqrt[]{ab}}:\dfrac{1}{\sqrt[]{a}-\sqrt[]{b}}=\dfrac{\sqrt[]{ab}\left(\sqrt[]{a}+\sqrt[]{b}\right)}{\sqrt[]{ab}}.\left(\sqrt[]{a}-\sqrt[]{b}\right)=a-b\)

a, \(VT=\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)

\(=\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{20}-2\right)}{2}\)

\(=\frac{\sqrt{5-2\sqrt{5}+1}\left(3+\sqrt{5}\right)\left(2\sqrt{5}-2\right)}{2}\)

\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)2\left(\sqrt{5}-1\right)}{2}\)

\(=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)

\(=18-6\sqrt{5}+6\sqrt{5}-10=8=VP\)

b, \(VT=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5-2\sqrt{5}\sqrt{3}+3}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)

\(=2\left(16-15\right)=2=VP\)

a) vì ab > 0 nên chia cả hai vế Bất đẳng thức cho \(\sqrt{ab}\) ta được

\(\sqrt{\dfrac{c\left(a-c\right)}{ab}}+\sqrt{\dfrac{c\left(b-c\right)}{ab}}\le1\)

Áp dụng Bất đẳng thức Cauchy cho hai số

\(\Rightarrow\sqrt{\dfrac{c}{b}\left(\dfrac{a-c}{a}\right)}+\sqrt{\dfrac{c}{a}\left(\dfrac{b-c}{b}\right)}\le\dfrac{1}{2}\left(\dfrac{c}{b}+\dfrac{a-c}{a}\right)+\dfrac{1}{2}\left(\dfrac{c}{a}+\dfrac{b-c}{b}\right)=1\)

vậy nên ta có đpcm

\(\frac{2005}{\sqrt{2006} }+\frac{2006}{\sqrt{2005} }>\sqrt{2005}+\sqrt{2006} \)

<=>\(2005\sqrt{2005}+2006\sqrt{2006}>2005\sqrt{2006}+2006\sqrt{2005} \)

<=>\(\sqrt{2006}<\sqrt{2005} \)

Bạn ghi sai đề thì phải giả thiết phải là \(x+y+z+\sqrt{xyz}=4\)

Khi đó suy ra \(4\left(x+y+z\right)+4\sqrt{xyz}=16\)

Ta có: \(x\left(4-y\right)\left(4-z\right)=x[16-4\left(y+z\right)+yz]=x[4\left(x+y+z\right)+4\sqrt{xyz}-4\left(y+z\right)+yz]\)

\(=x\left(4x+4\sqrt{xyz}+yz\right)=x\left(2\sqrt{x}+\sqrt{yz}\right)^2\)

\(\Rightarrow\sqrt{x\left(4-y\right)\left(4-z\right)}=\sqrt{x}\left(2\sqrt{x}+\sqrt{yz}\right)=2x+\sqrt{xyz}\)

tương tự \(\left\{{}\begin{matrix}\sqrt{y\left(4-z\right)\left(4-x\right)}=2y+\sqrt{xyz}\\\sqrt{z\left(4-x\right)\left(4-y\right)}=2z+\sqrt{xyz}\end{matrix}\right.\)

Cộng lại ta được VT\(=\) \(2\left(x+y+z+\sqrt{xyz}\right)+\sqrt{xyz}\) \(=8+\sqrt{xyz}\)(điều phải chứng minh)

 ta có:\(\frac{\left(x\sqrt{y}+y\sqrt{x}\right)\cdot\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=x-y\)

vậy.....

\(\frac{\left(x\sqrt{y}+y\sqrt{x}\right).\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)

\(=\frac{\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)

\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\)

\(=x-y\)( đpcm )

Cảm ơn bạn nhiều !!!

1) 

a) Ta có : \(\frac{x^2+5}{\sqrt{x^2+4}}=\frac{\left(x^2+4\right)+1}{\sqrt{x^2+4}}=\sqrt{x^2+4}+\frac{1}{\sqrt{x^2+4}}\). Đến đây áp dụng bđt \(a+\frac{1}{a}>2\)là ra nhé :)

b) Ta sẽ chứng minh bằng biến đổi tương đương : 

\(\sqrt{\left(a+c\right)\left(b+d\right)}\ge\sqrt{ab}+\sqrt{cd}\)

\(\Leftrightarrow\left(a+c\right)\left(b+d\right)\ge\left(\sqrt{ab}+\sqrt{cd}\right)^2\)

\(\Leftrightarrow ab+ad+bc+cd\ge ab+cd+2\sqrt{abcd}\)

\(\Leftrightarrow ad-2\sqrt{abcd}+bc\ge0\)

\(\Leftrightarrow\left(\sqrt{ad}-\sqrt{bc}\right)^2\ge0\)(luôn đúng)

Vì bđt cuối luôn đúng nên bđt ban đầu được chứng minh.

2) Mình làm tóm tắt thôi nhé , do đề dài...

a) \(\sqrt{2x+\sqrt{4x-1}}-\sqrt{2x-\sqrt{4x-1}}\)

\(=\frac{\sqrt{\left(4x-1\right)+2\sqrt{4x-1}+1}+\sqrt{\left(4x-1\right)-2\sqrt{4x-1}+1}}{\sqrt{2}}\)

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

b) \(\frac{x-y+3\sqrt{x}+3\sqrt{y}}{\sqrt{x}-\sqrt{y}+3}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+3\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}+3}\)

\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}+3\right)}{\sqrt{x}-\sqrt{y}+3}=\sqrt{x}+\sqrt{y}\)

c) Biến đổi  : \(\sqrt{x-2\sqrt{x-1}}=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}-1\right)^2}=\left|\sqrt{x-1}-1\right|\)

d) Biến đổi tương tự c) 

e) \(\sqrt{x+\sqrt{x^2-4}}.\sqrt{x-\sqrt{x^2-4}}=\sqrt{x^2-\left(x^2-4\right)}=\sqrt{4}=2\)