Bài 1:

a) Áp dụng BĐT Cô-si:

\(VT=a-1+\frac{1}{a-1}+1\ge2\sqrt{\frac{a-1}{a-1}}+1=2+1=3\)

Dấu "=" xảy ra \(\Leftrightarrow a=2\).

b) BĐT \(\Leftrightarrow a^2+2\ge2\sqrt{a^2+1}\)

\(\Leftrightarrow a^2+1-2\sqrt{a^2+1}+1\ge0\)

\(\Leftrightarrow\left(\sqrt{a^2+1}-1\right)^2\ge0\) ( LĐ )

Dấu "=" xảy ra \(\Leftrightarrow a=0\).

Bài 2: tương tự 1b.

Bài 3:

Do \(a,b,c\) dương nên ta có các BĐT:

\(\frac{a}{a+b+c}< \frac{a}{a+b}< \frac{a+c}{a+b+c}\)

Tương tự: \(\frac{b}{a+b+c}< \frac{b}{b+c}< \frac{b+a}{a+b+c};\frac{c}{a+b+c}< \frac{c}{c+a}< \frac{c+b}{a+b+c}\)

Cộng theo vế 3 BĐT:

\(\frac{a+b+c}{a+b+c}< \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{2\left(a+b+c\right)}{a+b+c}\)

\(\Leftrightarrow1< \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< 2\)( đpcm )

\(\frac{xy}{z}+\frac{yz}{x}\ge2y\) ; \(\frac{xy}{z}+\frac{zx}{y}\ge2x\); \(\frac{yz}{x}+\frac{zx}{y}\ge2z\)

Cộng vế với vế:

\(2\left(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\right)\ge2\left(x+y+z\right)\)

Dấu "=" xảy ra khi \(x=y=z\)

a. \(\)Áp dụng bất đẳng thức Côsi cho 2 số dương \(\frac{xy}{z}\) và \(\frac{yz}{x}\), ta có: \(\frac{xy}{z}+\frac{yz}{x}\ge2\sqrt{\frac{xy}{z}.\frac{yz}{x}}=2\sqrt{y^2}=2y\) (1)

Hoàn toàn tương tự: \(\)

\(\frac{yz}{x}+\frac{zx}{y}\ge2z\) và \(\frac{xy}{z}+\frac{zx}{y}\ge2x\) (2)

Từ (1) và (2) suy ra đpcm

\(\frac{xy\sqrt{z-1}+xz\sqrt{y-2}+yz\sqrt{x-3}}{xyz}\\ =\frac{xy\sqrt{z-1}}{xyz}+\frac{xz\sqrt{y-2}}{xyz}+\frac{yz\sqrt{x-3}}{xyz}\\ =\frac{\sqrt{z-1}}{z}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{x-3}}{x}\\ =\frac{2\sqrt{z-1}}{2z}+\frac{2\sqrt{2}\sqrt{y-2}}{2\sqrt{2}y}+\frac{2\sqrt{3}\sqrt{x-3}}{2\sqrt{3}x}\)

Áp dụng BDT Cô-si với 2 số không âm:

\(\Rightarrow\frac{2\sqrt{z-1}}{2z}+\frac{2\sqrt{2}\sqrt{y-2}}{2\sqrt{2}y}+\frac{2\sqrt{3}\sqrt{x-3}}{2\sqrt{3}x}\\ \le\frac{1+\left(z-1\right)}{2z}+\frac{2+\left(y-2\right)}{2\sqrt{2}y}+\frac{3+\left(x-3\right)}{2\sqrt{3}x}\\ =\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}=\frac{1}{2}+\frac{\sqrt{2}}{4}+\frac{\sqrt{3}}{6}\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}z-1=1\\y-2=2\\x-3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=2\\y=4\\x=6\end{matrix}\right.\)

Vậy.......

Đặt \(A=\frac{xy\sqrt{z-1}+xz\sqrt{y-2}+yz\sqrt{x-3}}{xyz}\)

\(\Rightarrow A=\frac{\sqrt{z-1}}{z}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{x-3}}{x}\)

\(\Rightarrow A=\frac{2.\sqrt{z-1}}{2z}+\frac{2.\sqrt{2}.\sqrt{y-2}}{2.\sqrt{2}.y}+\frac{2.\sqrt{3}.\sqrt{x-3}}{2.\sqrt{3}.x}\)\

\(\Rightarrow A\le\frac{z-1+1}{2z}+\frac{y-2+2}{2\sqrt{2}.y}+\frac{z-3+3}{2\sqrt{3}.x}\) ( ÁP DỤNG BĐT CÔ-SI )

\(\Rightarrow A\le\frac{z}{2z}+\frac{y}{2\sqrt{2}.y}+\frac{z}{2\sqrt{3}.z}\)

\(\Rightarrow A\le\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}=\frac{1}{2}+\frac{\sqrt{2}}{4}+\frac{\sqrt{3}}{6}\)

\(A=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\)

Áp dụng BĐT AM-GM ta có:

\(A\le\frac{1+x-1}{x}+\frac{2+y-2}{2y}+\frac{3+z-3}{3z}=1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}=1\\\sqrt{y-2}=2\\\sqrt{z-3}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x-1=1\\y-2=2\\z-3=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=4\\z=6\end{cases}}\)

Vậy \(A_{max}=\frac{11}{6}\Leftrightarrow\hept{\begin{cases}x=2\\y=4\\z=6\end{cases}}\)

Xin lỗi bạn. Bài đó mk lm sai rồi.

Sửa:

Áp dụng BĐT AM-GM ta có:

\(A=\frac{1.\sqrt{x-1}}{x}+\frac{\sqrt{2}.\sqrt{y-2}}{\sqrt{2}.y}+\frac{\sqrt{3}.\sqrt{z-3}}{\sqrt{3}.z}\le\frac{\frac{1+x-1}{2}}{x}+\frac{\frac{2+y-2}{2}}{\sqrt{2}.y}+\frac{\frac{3+z-3}{2}}{\sqrt{3}.z}=\frac{1}{2}+\frac{1}{2.\sqrt{2}}+\frac{1}{2.\sqrt{3}}\)\(=\frac{\sqrt{6}+\sqrt{3}+\sqrt{2}}{2.\sqrt{6}}\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}=1\\\sqrt{y-2}=\sqrt{2}\\\sqrt{z-3}=\sqrt{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}x-1=1\\y-2=2\\z-3=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=4\\z=6\end{cases}}\)

Vậy \(A_{max}=\frac{\sqrt{6}+\sqrt{2}+\sqrt{3}}{2.\sqrt{6}}\)\(\Leftrightarrow\hept{\begin{cases}x=2\\y=4\\z=6\end{cases}}\)

Ta có: \(\frac{1+ab}{1+a^2}+\frac{1+ab}{1+b^2}=\left(1+ab\right)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}\right)\)

mà \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{4}{2+a^2+b^2}\)( Áp dụng BĐT phụ \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\))

Mặt khác: \(a^2+b^2\ge2ab\)

=> \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{4}{2+2ab}=\frac{2}{1+ab}\)

=> \(\left(1+ab\right)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}\right)\ge\left(1+ab\right)\left(\frac{2}{1+ab}\right)=2\)(đpcm)

3. a) \(A=x+\frac{1}{x-1}=x-1+\frac{1}{x-1}+1\ge2\sqrt{\left(x-1\right)\cdot\frac{1}{x-1}}+1=3\)

Dấu "=" \(\Leftrightarrow x-1=\frac{1}{x-1}\Leftrightarrow x=2\)

Min \(A=3\Leftrightarrow x=2\)

b) \(B=\frac{4}{x}+\frac{1}{4y}=\frac{4}{x}+4x+\frac{1}{4y}+4y\cdot-4\left(x+y\right)\)

\(\ge2\sqrt{\frac{4}{x}\cdot4x}+2\sqrt{\frac{1}{4y}\cdot4y}-4\cdot\frac{5}{4}=5\)

Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}\frac{4}{x}=4x\\\frac{1}{4y}=4y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\frac{1}{4}\end{matrix}\right.\)

Min \(B=5\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\frac{1}{4}\end{matrix}\right.\)

4. Chắc đề là tìm min???

\(C=a+b+\frac{1}{a}+\frac{1}{b}\ge a+b+\frac{4}{a+b}=a+b+\frac{1}{a+b}+\frac{3}{a+b}\)

\(\ge2\sqrt{\left(a+b\right)\cdot\frac{1}{a+b}}+\frac{3}{1}=5\)

Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}a=b\\a+b=\frac{1}{a+b}\\a+b=1\end{matrix}\right.\Leftrightarrow a=b=\frac{1}{2}\)

Min \(C=5\Leftrightarrow a=b=\frac{1}{2}\)

1. Áp dụng bđt \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) ta có:

\(\left(\frac{1}{p-a}+\frac{1}{p-b}\right)+\left(\frac{1}{p-b}+\frac{1}{p-c}\right)+\left(\frac{1}{p-c}+\frac{1}{p-a}\right)\)

\(\ge\frac{4}{2p-a-b}+\frac{4}{2p-b-c}+\frac{4}{2p-a-c}\) \(=\frac{4}{c}+\frac{4}{a}+\frac{4}{b}\)

\(\Rightarrow\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Dấu "=" \(\Leftrightarrow a=b=c\)

2. Áp dụng bđt Cauchy ta có :

\(a\sqrt{b-1}=a\sqrt{\left(b-1\right)\cdot1}\le a\cdot\frac{b-1+1}{2}=\frac{ab}{2}\) . Dấu "=" \(\Leftrightarrow b-1=1\Leftrightarrow b=2\)

+ Tương tự : \(b\sqrt{a-1}\le\frac{ab}{2}\). Dấu "=" \(\Leftrightarrow a=2\)

Do đó: \(a\sqrt{b-1}+b\sqrt{a-1}\le ab\). Dấu "=" \(\Leftrightarrow a=b=2\)

5. Tìm min???

Áp dụng bđt Cauchy ta có:

\(a^3+b^3+1\ge3\sqrt[3]{a^3\cdot b^3\cdot1}=3ab\) . Dấu "=" \(\Leftrightarrow a=b=1\)

+ Tương tự : \(a^3+c^3+1\ge3ac\). Dấu "=" \(\Leftrightarrow a=c=1\)

\(b^3+c^3+1\ge3bc\). Dấu "=" \(\Leftrightarrow b=c=1\)

Do đó : \(2D+3\ge3\left(ab+bc+ca\right)\Rightarrow D\ge3\)

Min \(D=3\Leftrightarrow a=b=c=1\)