Câu 1 : áp dụng BĐT SVAC ta có \(A\ge\frac{(a+b+c)^2}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c}}=\frac{1.\sqrt{2a+2b+2c}}{\sqrt{2.}(\sqrt{b+c}+\sqrt{a+b}+\sqrt{a+c})}\)

mặt khác lại có \(\frac{\sqrt{2a+2b+2c}}{\sqrt{2}.(\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c})}\ge\frac{\sqrt{(\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c})^2}}{\sqrt{2}.\sqrt{3}.(\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c})}=\frac{1}{\sqrt{6}}\)theo bđt svac

\(\Rightarrow A\ge\frac{1}{\sqrt{6}}\)dấu bằng xảy ra tại a=b=c=\(\frac{1}{3}\)

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

Áp dụng BĐT Mincopxki:

\(P\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)

\(\ge\sqrt{\left(a+b+c\right)^2+\dfrac{81}{\left(a+b+c\right)^2}}\)

\(\ge\sqrt{\left(a+b+c\right)^2+\dfrac{81}{16\left(a+b+c\right)^2}+\dfrac{1215}{16\left(a+b+c\right)^2}}\)

\(\ge\sqrt{2\sqrt{\left(a+b+c\right)^2\cdot\dfrac{81}{16\left(a+b+c\right)^2}}+\dfrac{1215}{16\cdot\left(\dfrac{3}{2}\right)^2}}\)

\(=\dfrac{3\sqrt{17}}{2}\)

\("="\Leftrightarrow a=b=c=\dfrac{1}{2}\)

Cách khác :)

Áp dụng bất đẳng thức Bunhiacopxki :

\(\left(1+16\right)\left(a^2+\frac{1}{b^2}\right)\ge\left(a+\frac{4}{b}\right)^2\)

\(\Rightarrow\sqrt{17}\cdot\sqrt{a^2+\frac{1}{b^2}}\ge a+\frac{4}{b}\)

Tương tự : \(\sqrt{17}\cdot\sqrt{b^2+\frac{1}{c^2}}\ge b+\frac{4}{c};\sqrt{17}\cdot\sqrt{c^2+\frac{1}{a^2}}\ge c+\frac{4}{a}\)

Cộng theo vế của 3 bất đẳng thức :

\(\sqrt{17}\cdot\left(\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{c^2}}+\sqrt{c^2+\frac{1}{a^2}}\right)\ge\left(a+b+c\right)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Leftrightarrow\sqrt{17}\cdot P\ge a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)

Áp dụng bất đẳng thức Cô-si:

Xét \(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)

\(=16a+\frac{4}{a}+16b+\frac{4}{b}+16c+\frac{4}{c}-15a-15b-15c\)

\(\ge2\sqrt{\frac{16\cdot4a}{a}}+2\sqrt{\frac{16\cdot4b}{b}}+2\sqrt{\frac{16\cdot4c}{c}}-15\left(a+b+c\right)\)

\(=16\cdot3-15\cdot\frac{3}{2}=\frac{51}{2}\)

Ta có : \(\sqrt{17}\cdot P\ge\frac{51}{2}\)

\(\Leftrightarrow P\ge\frac{3\sqrt{17}}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{2}\)

Em làm thử nhé!

Bài 1: \(A=\left[\frac{a^2}{b-1}+4\left(b-1\right)\right]+\left[\frac{b^2}{a-1}+4\left(a-1\right)\right]-4\left(a+b\right)+8\)

Cauchy vào là ra rồi ạ;)

Bài 2: Em chịu

2) Có: \(\sqrt{ab}\le\frac{a+b}{2}=1\); \(\sqrt{a}+\sqrt{b}=\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2}\le\sqrt{2\left(a+b\right)}=2\)

\(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}=\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{ab}}\ge\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3=\frac{a^2}{\sqrt{a}}+\frac{b^2}{\sqrt{b}}\)

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

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=1\)

ko khó nhưng mà bn đăng từng câu 1 hộ mk mk giải giúp cho

gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)

=> Thay vào thì     \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)

\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)

Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào

=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)

=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)

=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\) 

Đặt: \(\sqrt{a}=x;\sqrt{b}=y;\sqrt{c}=z\)

=>     \(P=\frac{xy}{z^2+3xy}+\frac{yz}{x^2+3yz}+\frac{zx}{y^2+3zx}\)

=>     \(3P=\frac{3xy}{z^2+3xy}+\frac{3yz}{x^2+3yz}+\frac{3zx}{y^2+3zx}=1-\frac{z^2}{z^2+3xy}+1-\frac{x^2}{x^2+3yz}+1-\frac{y^2}{y^2+3zx}\)

Ta sẽ CM: \(3P\le\frac{9}{4}\)<=> Cần CM: \(\frac{x^2}{x^2+3yz}+\frac{y^2}{y^2+3zx}+\frac{z^2}{z^2+3xy}\ge\frac{3}{4}\)

Có:    \(VT\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\)

Ta sẽ CM: \(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\ge\frac{3}{4}\)

<=> \(4\left(x+y+z\right)^2\ge3\left(x^2+y^2+z^2\right)+9\left(xy+yz+zx\right)\)

<=> \(4\left(x^2+y^2+z^2\right)+8\left(xy+yz+zx\right)\ge3\left(x^2+y^2+z^2\right)+9\left(xy+yz+zx\right)\)

<=> \(x^2+y^2+z^2\ge xy+yz+zx\)

Mà đây lại là 1 BĐT luôn đúng => \(3P\le\frac{9}{4}\)=> \(P\le\frac{3}{4}\)

Vậy P max \(=\frac{3}{4}\)<=> \(a=b=c\)