Kurosaki Akatsu giải thế thì đề bài cho  \(b^2+c^2\le a^2\)  để làm gì?

Áp dụng bất đẳng thức AM-GM ta có :

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

\(P=\frac{b^2}{a^2}+\frac{c^2}{a^2}+\frac{a^2}{b^2}+\frac{a^2}{c^2}\ge4.\sqrt[4]{\frac{b^2}{a^2}.\frac{c^2}{a^2}.\frac{a^2}{b^2}.\frac{a^2}{c^2}}=4.1=4\)

=> \(Min_P=4\)

Với a, b, c thực dương áp dụng BĐT Cô-si ta có:

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

\(\ge2\sqrt{\frac{b^2}{a^2}.\frac{c^2}{a^2}}+2\sqrt{\frac{a^2}{b^2}.\frac{a^2}{c^2}}=2\left(\frac{bc}{a^2}+\frac{a^2}{bc}\right)\)

\(=2\left[\left(\frac{bc}{a^2}+\frac{a^2}{4bc}\right)+\frac{3a^2}{4bc}\right]\ge2\left(2.\sqrt{\frac{bc}{a^2}.\frac{a^2}{4bc}}+\frac{3\left(b^2+c^2\right)}{4bc}\right)\)   (vì  \(a^2\ge b^2+c^2\))

\(=2\left(2\sqrt{\frac{1}{4}}+\frac{3.2bc}{4bc}\right)\) (vì  \(b^2+c^2\ge2bc\))

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

Vậy  P_min = 5

Đẳng thức xảy ra  \(\Leftrightarrow\)  \(\hept{\begin{cases}a^2=b^2+c^2\\b=c\end{cases}}\)

Ta có: \(\frac{1+3a}{1+b^2}=\left(1+3a\right).\frac{1}{1+b^2}=\left(1+3a\right)\left(1-\frac{b^2}{1+b^2}\right)\)

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

\(=3a+1-\frac{b}{2}-\frac{3ab}{2}\)(1)

Tương tự ta có: \(\frac{1+3b}{1+c^2}=3b+1-\frac{c}{2}-\frac{3bc}{2}\)(2); \(\frac{1+3c}{1+a^2}=3c+1-\frac{a}{2}-\frac{3ca}{2}\)(3)

Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(\frac{1+3a}{1+b^2}+\frac{1+3b}{1+c^2}+\frac{1+3c}{1+a^2}\)\(\ge3\left(a+b+c\right)-\frac{a+b+c}{2}-\frac{3\left(ab+bc+ca\right)}{2}+3\)

\(=\frac{5\left(a+b+c\right)}{2}-\frac{3\left(ab+bc+ca\right)}{2}+3\)

\(\ge\frac{5.\sqrt{3\left(ab+bc+ca\right)}}{2}-\frac{3.3}{2}+3=\frac{15}{2}-\frac{9}{2}+3=6\)

Đẳng thức xảy ra khi a = b = c = 1

Bài toán số 41 có 2 cách làm, tôi làm cách thứ 2

Đặt \(Q=\sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{x+z}}+\sqrt{\frac{z}{x+y}}\)\(\Rightarrow Q^2=\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}+2\left(\sqrt{\frac{xy}{\left(y+z\right)\left(x+z\right)}}+\sqrt{\frac{yz}{\left(x+z\right)\left(y+z\right)}}+\sqrt{\frac{xz}{\left(x+y\right)\left(y+z\right)}}\right)\)ta thấy rằng \(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{1}{4}\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)\left(xy+yz+zx\right)\)

\(=\frac{x^2+y^2+z^2}{4}+\frac{xyz}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\ge\frac{x^2+y^2+z^2}{4}\)

Áp dụng bất đẳng thức AM-GM ta có \(\sqrt{\frac{yx}{\left(z+x\right)\left(x+y\right)}}\ge\frac{2yx}{2\sqrt{\left(xy+yz\right)\left(yz+yx\right)}}\ge\frac{2xy}{2xy+yz+xz}\ge\frac{2xy}{2\left(xy+yz+zx\right)}=\frac{xy}{xy+yz+zx}\)

Tương tự ta có \(\hept{\begin{cases}\sqrt{\frac{yz}{\left(z+x\right)\left(z+y\right)}}\ge\frac{yz}{xy+yz+zx}\\\sqrt{\frac{xz}{\left(x+y\right)\left(y+z\right)}}\ge\frac{xz}{xy+yz+zx}\end{cases}}\)

\(\Rightarrow\sqrt{\frac{xy}{\left(y+z\right)\left(z+x\right)}}+\sqrt{\frac{yz}{\left(z+x\right)\left(x+y\right)}}+\sqrt{\frac{zx}{\left(x+y\right)\left(y+z\right)}}\ge1\)nên \(Q\ge\sqrt{\frac{x^2+y^2+z^2}{4}+2}\)

\(\Rightarrow Q\ge\sqrt{\frac{x^2+y^2+z^2}{2}+4}+\frac{4}{\sqrt{x^2+y^2+z^2}}\)

Đặt \(t=\sqrt{x^2+y^2+z^2}\Rightarrow t\ge\sqrt{xy+yz+zx}=2\)

Xét hàm số g(t)=\(\sqrt{\frac{t^2}{2}+4}+\frac{4}{t}\left(t\ge2\right)\)khi đó ta có 

\(g'\left(t\right)=\frac{t}{2\sqrt{\frac{t^2}{2}+4}}-\frac{4}{t^2};g'\left(t\right)=0\Leftrightarrow t^6-32t^2-256=0\Leftrightarrow t=2\sqrt{2}\)

Lập bảng biến thiên ta có min[2;\(+\infty\)) \(g\left(t\right)=g\left(2\sqrt{2}\right)=3\sqrt{2}\)

Hay minS=\(3\sqrt{2}\)<=> a=c=1; b=2

Đặt a=xc; b=cy (x;y >=1)

\(\sqrt{x-1}+\sqrt{y-1}=\sqrt{xy}\Leftrightarrow x+y-2+2\sqrt{\left(x-1\right)\left(y-1\right)}=xy\)

\(\Leftrightarrow xy-x-y+1-2\sqrt{\left(x-1\right)\left(y-1\right)}+1=0\)

\(\Leftrightarrow\left(\sqrt{\left(x-1\right)\left(y-1\right)}-1\right)^2=0\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(y-1\right)}=1\Leftrightarrow xy=x+y\ge2\sqrt{xy}\Rightarrow xy\ge4\)

Biểu thức P được viết lại như sau

\(P=\frac{x}{y+1}+\frac{y}{x+1}+\frac{1}{x+y}+\frac{1}{x^2+y^2}=\frac{x^2}{xy+x}+\frac{y^2}{xy+y}+\frac{1}{x^2+y^2}+\frac{1}{\left(x+y\right)^2-2xy}\)

\(P\ge\frac{\left(x+y\right)^2}{2xy+x+y}+\frac{1}{x+y}+\frac{1}{\left(x+y\right)^2-2xy}=\frac{xy}{3}+\frac{1}{xy}+\frac{1}{x^2y^2-2xy}=\frac{x^3y^3-2x^2y^2+3xy-3}{3\left(x^2y^2-2xy\right)}\)

Đặt t=xy với t>=4

Xét hàm số \(f\left(t\right)=\frac{t^3-2t^2+3t-3}{t^2-2t}\left(t\ge4\right)\)

Ta có \(f'\left(t\right)=\frac{t^4-4t^3+t^2+6t-6}{\left(t^2-2t\right)^2}=\frac{t^3\left(t-4\right)+6\left(t-4\right)+18}{\left(t^2-2t\right)^2}>0\forall t\ge4\)

Lập bảng biến thiên ta có \(minf\left(t\right)=f\left(4\right)=\frac{41}{8}\)

Vậy \(minP=\frac{41}{24}\)khi x=y=z=2 hay a=b=2c

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

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

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

Ta có \(\sqrt{1+8a^3}=\sqrt{\left(1+2a\right)\left(1-2a+4a^2\right)}\le\frac{1+2a+1-2a+4a^2}{2}=1+2a^2\)(BĐT AM-GM)

Tương tự cho \(\sqrt{1+8b^2};\sqrt{1+8c^2}\)ta được \(P\ge\frac{1}{1+2a^2}+\frac{1}{1+2b^2}+\frac{1}{1+2c^2}\)

Mặt khác \(\frac{1}{1+2a^2}=\frac{1}{1+2a^2}+\frac{1+2a^2}{9}-\frac{1+2a^2}{9}\ge2\sqrt{\frac{1}{1+2a^2}\cdot\frac{1+2a^2}{9}}-\frac{2}{9}a^2-\frac{1}{9}=\frac{5-2a^2}{9}\)

Khi đó: \(P\ge\frac{5-2a^2}{9}-\frac{5-2b^2}{9}-\frac{5-2c^2}{9}\) \(=\frac{15-2\left(a^2+b^2+c^2\right)}{9}=\frac{15-2\cdot3}{9}=1\)

Vậy Min P=1

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a^2+b^2+c^2=3\\1+2a=1-2a+4a^2\\\frac{1}{1+2a^2}=\frac{1+2a^2}{9}\end{cases}}\)và vai trò a,b,c như nhau hay (a,b,c)=(1,1,1)

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!