+)\(\frac{3}{4}\ge a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\Leftrightarrow\frac{1}{8}\ge abc\)

+) \(P=8abc+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left(32abc+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)-24abc\)

\(\ge4\sqrt[4]{\frac{32}{abc}}-24abc\ge4\sqrt[4]{\frac{32}{\frac{1}{8}}}-3=16-3=13\)

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

để biểu thức cho đơn giản , ta đặt x=a+1,y=b+1,z=c+1(x,y,z>0)

thì giả thiết thành \(\frac{1}{x+1}+\frac{3}{y+3}\le\frac{z}{z+2}\) .Tìm min xyz 

Áp dụng bất đẳng thức cauchy:\(\frac{z}{z+2}\ge\frac{1}{x+1}+\frac{3}{y+3}\ge2\sqrt{\frac{3}{\left(x+1\right)\left(y+3\right)}}\)(1)

từ giả thiết :\(\frac{1}{x+1}\le\frac{z}{z+2}-\frac{3}{y+3}\Leftrightarrow1-\frac{1}{x+1}\ge1-\frac{z}{z+2}+\frac{3}{y+3}\)

\(\Leftrightarrow\frac{x}{x+1}\ge\frac{2}{z+2}+\frac{3}{y+3}\)

Áp dụng bất đẳng thức cauchy 1 lần nữa: \(\frac{x}{x+1}\ge\frac{2}{z+2}+\frac{3}{y+3}\ge2\sqrt{\frac{6}{\left(z+2\right)\left(y+3\right)}}\)(2)

tương tự ta cũng có: \(\frac{y}{y+3}\ge2\sqrt{\frac{2}{\left(z+2\right)\left(x+1\right)}}\)(3),

cả 2 vế các bất đẳng thức (1),(2)và (3) đều dương, nhân vế với vế: 

\(\frac{xyz}{\left(x+1\right)\left(y+3\right)\left(z+2\right)}\ge\frac{8.6}{\left(x+1\right)\left(z+2\right)\left(y+3\right)}\)

\(\Leftrightarrow xyz\ge48\)

Dấu = xảy ra khi x=2,y=6,z=4 hay a=1,b=5,z=3

1a

\(A=\frac{3}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^4+b^4}{2}\ge\frac{6}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^2+b^2\right)^2}{2}}{2}\)

\(\ge10+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{4}=10+\frac{1}{16}=\frac{161}{16}\)

Dau '=' xay ra khi \(a=b=\frac{1}{2}\)

Vay \(A_{min}=\frac{161}{16}\)

1b.\(B=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^8+b^8}{4}\ge\frac{2}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^4+b^4\right)^2}{2}}{4}\)

\(\ge6+\frac{\left[\frac{\left(a^2+b^2\right)^2}{2}\right]^2}{8}\ge6+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{32}=6+\frac{1}{128}=\frac{769}{128}\)

Dau '=' xay ra khi \(a=b=\frac{1}{2}\)

Vay \(B_{min}=\frac{769}{128}\)khi \(a=b=\frac{1}{2}\)

Bài 2 Dùng Cauchy-Schwarz dạng Engel là ra:D

Bài 3:Đừng vội dùng Cauchy-Schwarz dạng Engel ngay kẻo bị phức tạp:v Thay vào đó hãy khai triển nó ra:

\(A=x^2+y^2+2\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{1}{x^2}+\frac{1}{y^2}\)

\(\ge4+2.2+\frac{4}{x^2+y^2}=4+4+1=9\)

Đẳng thức xảy ra khi \(x=y=\sqrt{2}\)

Bài 4: Dùng Cauchy or Bunhiacopxki là ok!

\(GT\Rightarrow\)\(\frac{1}{a+2}+\frac{3}{b+4}\leq1-\frac{2}{c+3}\)

Áp dụng BĐT AM-GM ta có:

\(1-\frac{2}{c+3}\geq\frac{1}{a+2}+\frac{3}{b+4}\geq2\sqrt{\frac{3}{(a+2)(b+4)}}\)

Tương tự ta có:

\(1-\frac{1}{a+2}\geq\frac{3}{b+4}+\frac{2}{c+3}\geq2\sqrt{\frac{6}{(c+3)(b+4)}}\)

\(1-\frac{3}{b+4}\geq\frac{1}{a+2}+\frac{2}{c+3}\geq2\sqrt{\frac{6}{(c+3)(a+2)}}\)

Nhân theo vế ta được: \((1-\frac{2}{c+3})(1-\frac{1}{a+2})(1-\frac{3}{b+4})\geq \frac{48}{(a+2)(b+4)(c+3)}\)

\(\Leftrightarrow (\frac{c+1}{c+3})(\frac{a+1}{a+2})(\frac{b+1}{b+4})\geq\frac{48}{(a+2)(b+4)(c+3)}\)

\(\Leftrightarrow(a+1)(b+1)(c+1)\geq48\)

Dấu "=" xảy ra khi \(a=1;c=3;b=5\)

\(Gt\Leftrightarrow 1-\frac{1}{a+2}+1-\frac{3}{b+4}+\frac{c+1}{c+3}\geq 2\\\Leftrightarrow \frac{a+1}{a+2}+\frac{b+1}{b+4}+\frac{c+1}{c+3}\geq 2\)

Đặt \((a+1;b+1;c+1)\rightarrow (x;y;z)\), vậy cần tìm GTNN của \(Q=xyz\)

Ta có: \(\frac{x}{x+1}+\frac{y}{y+3}+\frac{z}{z+2}\geq 2\)

Áp dụng BĐT AM-GM ta có:

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

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

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

Nhân theo vế ta có:\(\frac{xyz}{\left(x+1\right)\left(y+3\right)\left(z+2\right)}\ge\frac{48}{\left(x+1\right)\left(y+3\right)\left(z+2\right)}\Leftrightarrow Q\ge48\)

Dấu "=" xảy ra khi \(\Leftrightarrow \left\{\begin{matrix} \frac{1}{x+1}=\frac{3}{y+3}=\frac{2}{z+2} & & \\ \frac{1}{a+2}+\frac{3}{b+4}=\frac{c+1}{c+3} & & \end{matrix}\right.\)\(\Leftrightarrow a=1;b=5;c=3\)

mk còn hơi nhiều cách giải cần tham khảo thì pm

Bài 1:

Áp dụng BĐT Bunhiacopxky ta có:

\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x+y+z)\geq (1+1+1)^2\)

\(\Leftrightarrow A.1\geq 9\Leftrightarrow A\geq 9\)

Vậy GTNN của $A$ là $9$. Giá trị này đạt được tại $x=y=z=\frac{1}{3}$

Bài 2:

Hoàn toàn tương tự bài 1

$S(a+b+c)\geq (1+1+1)^2$ theo BĐT Bunhiacopxky

$\Leftrightarrow S.3\geq 9\Rightarrow S\geq 3$

Vậy GTNN của $S$ là $3$ khi $a=b=c=1$

Bài 3:

Áp dụng BĐT Bunhiacopxky như các bài trên ta có:

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$

Mà $0< x+y+z\leq 6$ nên $\frac{9}{x+y+z}\geq \frac{9}{6}=\frac{3}{2}$

Do đó $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{3}{2}$ (đpcm)

Dấu "=" xảy ra khi $x=y=z=2$

Bài 4:

Áp dụng BĐT Cô-si cho các số dương ta có:

$a^4+b^4+c^4+d^4\geq 4\sqrt[4]{a^4b^4c^4d^4}=4abcd$ (đpcm)

Dấu "=" xảy ra khi $a=b=c=d>0$

\(\frac{c+1}{c+3}\ge\frac{1}{a+2}+\frac{3}{b+4}\ge2\sqrt[]{\frac{3}{\left(a+2\right)\left(b+4\right)}}\) (1)

\(\frac{1}{a+2}+\frac{3}{b+4}\le\frac{c+3-2}{c+3}=1-\frac{2}{c+3}\Rightarrow1-\frac{1}{a+2}\ge\frac{3}{b+4}+\frac{2}{c+3}\)

\(\Rightarrow\frac{a+1}{a+2}\ge\frac{3}{b+4}+\frac{2}{c+3}\ge2\sqrt{\frac{6}{\left(b+4\right)\left(c+3\right)}}\) (2)

\(\frac{1}{a+2}+\frac{3}{b+4}\le1-\frac{2}{c+3}\Rightarrow1-\frac{3}{b+4}\ge\frac{1}{a+2}+\frac{2}{c+3}\)

\(\Rightarrow\frac{b+1}{b+4}\ge\frac{1}{a+2}+\frac{2}{c+3}\ge2\sqrt{\frac{2}{\left(a+2\right)\left(c+3\right)}}\) (3)

Nhân vế với vế (1);(2);(3):

\(\frac{\left(a+1\right)\left(b+1\right)\left(c+1\right)}{\left(a+2\right)\left(b+4\right)\left(c+3\right)}\ge8\sqrt{\frac{36}{\left(a+2\right)^2\left(b+4\right)^2\left(c+3\right)^2}}=\frac{48}{\left(a+2\right)\left(b+4\right)\left(c+3\right)}\)

\(\Rightarrow Q\ge48\Rightarrow Q_{min}=48\) khi \(\left(a;b;c\right)=\left(1;5;3\right)\)

\(P=\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}+2-2=\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-2\)

\(=\left(\frac{a^2}{b^3}+\frac{1}{a}+\frac{1}{a}\right)+\left(\frac{b^2}{c^3}+\frac{1}{b}+\frac{1}{b}\right)+\left(\frac{c^2}{a^3}+\frac{1}{c}+\frac{1}{c}\right)-2\)

Áp dụng BĐT AM-GM cho 3 số dương: 

\(\frac{a^2}{b^3}+\frac{1}{a}+\frac{1}{a}\ge3\sqrt[3]{\frac{a^2}{b^3}.\frac{1}{a}.\frac{1}{a}}=\frac{3}{b}\)

\(\frac{b^2}{c^3}+\frac{1}{b}+\frac{1}{b}\ge3\sqrt[3]{\frac{b^2}{c^3}.\frac{1}{b}.\frac{1}{b}}=\frac{3}{c}\)

\(\frac{c^2}{a^3}+\frac{1}{c}+\frac{1}{c}\ge3\sqrt[3]{\frac{c^2}{a^3}.\frac{1}{c}.\frac{1}{c}}=\frac{3}{a}\)

\(\Rightarrow P\ge\frac{3}{b}+\frac{3}{c}+\frac{3}{a}-2=3-2=1\)

Dấu "=" xảy ra khi \(a=b=c=3\)

Đặt \(\frac{1}{a}=x,\frac{1}{b}=y,\frac{1}{c}=z\) thì

\(\Rightarrow\hept{\begin{cases}x+y+z=1\\P=\frac{y^3}{x^2}+\frac{z^3}{y^2}+\frac{x^3}{z^2}\end{cases}}\)

Ta có:

\(\frac{x^3}{z^2}+z+z\ge3x,\frac{y^3}{x^2}+x+x\ge3y,\frac{z^3}{y^2}+y+y\ge3z\)

\(\Rightarrow\frac{x^3}{z^2}\ge3x-2z,\frac{y^3}{x^2}\ge3y-2x,\frac{z^3}{y^2}\ge3z-2y\)

\(\Rightarrow P\ge3x-2z+3y-2x+3z-2y=x+y+z=1\)

Ta co:

\(1=a+b+c\ge3\sqrt[3]{abc}\Rightarrow abc\le\frac{1}{27}\)

Dat \(P=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{3}{\sqrt[3]{a^2b^2c^2}}\ge\frac{3}{\frac{1}{9}}=27\)

Dau '=' xay ra khi \(a=b=c=\frac{1}{3}\)

Vay \(P_{min}=27\)khi \(a=b=c=\frac{1}{3}\)

còn cách khác k bạn

Theo yêu cầu của bạn cách kh́ác :))

Ta co:

\(\frac{1}{a^2}+9\ge\frac{6}{a}\)

Tuong tu:\(\frac{1}{b^2}+9\ge\frac{6}{b};\frac{1}{c^2}+9\ge\frac{6}{c}\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+27\ge6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\frac{54}{a+b+c}=54\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge54-27=27\)

Dau '=' xay ra khi \(a=b=c=\frac{1}{3}\)

Vay \(P_{min}=27\)khi \(a=b=c=\frac{1}{3}\)