a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)

Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2

b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)

Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)

Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)

Áp dụng bđt bunhiacopxki, ta có:

\(\left(x^2+\frac{1}{x^2}\right)\left(1+16\right)\ge\left(x+\frac{4}{x}\right)^2\) => \(x^2+\frac{1}{x^2}\ge\frac{\left(x+\frac{4}{x}\right)^2}{17}\)

=> \(\sqrt{x^2+\frac{1}{x^2}}\ge\frac{x+\frac{4}{x}}{\sqrt{17}}=\frac{x}{\sqrt{17}}+\frac{4}{x\sqrt{17}}\)

CMTT: \(\sqrt{y^2+\frac{1}{y^2}}\ge\frac{y}{\sqrt{17}}+\frac{4}{\sqrt{17}y}\)

\(\sqrt{z^2+\frac{1}{z^2}}\ge\frac{z}{\sqrt{17}}+\frac{4}{\sqrt{17}z}\)

=> A \(\ge\frac{x+y+z}{\sqrt{17}}+\frac{4}{\sqrt{17}}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{x+y+z}{\sqrt{17}}+\frac{36}{\sqrt{17}\left(x+y+z\right)}\)(bđt: 1/a + 1/b + 1/c > = 9/(a+b+c)

=> A \(\ge\frac{16\left(x+y+z\right)}{\sqrt{17}}+\frac{36}{\sqrt{17}\left(x+y+z\right)}-\frac{15\left(x+y+z\right)}{\sqrt{17}}\)

A \(\ge2\sqrt{\frac{16\left(x+y+z\right)}{\sqrt{17}}\cdot\frac{36}{\sqrt{17}\left(x+y+z\right)}}-\frac{15\cdot\frac{3}{2}}{\sqrt{17}}\)(Bđt cosi + bđt: x + y + z < = 3/2)

A \(\ge\frac{48}{\sqrt{17}}-\frac{45}{2\sqrt{17}}=\frac{3\sqrt{17}}{2}\)

Dấu "=" xảy ra <=> x = y= z = 1/2

Vậy MinA = \(\frac{3\sqrt{17}}{2}\) <=> x = y = z = 1/2

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

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

\(=2+\left(x^2y^2+\frac{1}{256x^2y^2}\right)+\frac{255}{256x^2y^2}\)

Áp dụng BĐT Cauchy - Schwar cho 2 số không âm, ta được:

\(x^2y^2+\frac{1}{256x^2y^2}\ge2\sqrt{\frac{x^2y^2}{256x^2y^2}}=\frac{1}{8}\)

C/m được BĐT phụ: \(1=\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow16x^2y^2\le1\Leftrightarrow256x^2y^2\le16\Leftrightarrow\frac{255}{256x^2y^2}\ge\frac{255}{16}\)

\(\Rightarrow M\ge2+\frac{1}{8}+\frac{255}{16}=\frac{289}{16}\)

(Dấu "="\(\Leftrightarrow\hept{\begin{cases}x^2y^2=\frac{1}{256x^2y^2}\\x-y=0\end{cases}}\Leftrightarrow x=y=\frac{1}{2}\))

\(\frac{16}{3x+3y+2z}=\frac{16}{\left(x+y\right)+\left(y+z\right)+\left(z+x\right)+\left(x+y\right)1}\le\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\)

Tương tự \(\frac{16}{3x+2y+3z}\le\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+z}\)

\(\frac{16}{2x+3y+3z}\le\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{y+z}\)

Cộng vế theo vế ta có:

\(16\left(\frac{1}{3x+2y+3z}+\frac{1}{3x+3y+2z}+\frac{1}{2x+3y+3z}\right)\le4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=24\)

\(\Rightarrow\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\le\frac{3}{2}\left(đpcm\right)\)

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b) \(\frac{1}{3x+3y+2z}=\frac{1}{\left(x+y\right)+\left(y+z\right)+\left(x+z\right)+\left(x+y\right)}\le\frac{1}{16}\left(\frac{2}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)

Tương tự hai bđt còn lại và cộng theo vế thu được: \(VT\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{6}{4}=\frac{3}{2}\)

Đẳng thức xảy ra khi \(x=y=z=\frac{1}{4}\)

\(P=\sum\frac{x^2\left(y+z\right)}{yz}\ge\sum\frac{4x^2\left(y+z\right)}{\left(y+z\right)^2}=\sum\frac{4x^2}{y+z}\ge\frac{4\left(x+y+z\right)^2}{y+z+z+x+x+y}=2\left(x+y+z\right)=2\)

\(P_{min}=2\) khi \(x=y=z=\frac{1}{3}\)

Câu 2 có dương không nhỉ? Không dương thì không làm được

\(A=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\ge\frac{4}{x^2+y^2+2xy}+\frac{2}{\left(x+y\right)^2}=\frac{6}{\left(x+y\right)^2}\ge6\)

\(A_{min}=6\) khi \(x=y=\frac{1}{2}\)

1) \(P\ge\frac{x^2.2\sqrt{yz}}{yz}+\frac{y^2.2\sqrt{zx}}{zx}+\frac{z^2.2\sqrt{xy}}{xy}=\frac{2x^2}{\sqrt{yz}}+\frac{2y^2}{\sqrt{zx}}+\frac{2z^2}{\sqrt{xy}}\ge4\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)=4\left\{\left[\frac{x^2}{y+z}+\frac{1}{4}\left(y+z\right)\right]+\left[\frac{y^2}{z+x}+\frac{1}{4}\left(z+x\right)\right]+\left[\frac{z^2}{x+y}+\frac{1}{4}\left(x+y\right)\right]\right\}-2\left(x+y+z\right)\ge4\left(x+y+z\right)-2\left(x+y+z\right)=2\)

Dấu "=" xảy ra <=> \(x=y=z=\frac{1}{3}\)

2) \(A=\left[\frac{1}{x^2+y^2}+4\left(x^2+y^2\right)\right]+\left(\frac{1}{xy}+16xy\right)-4\left(x+y\right)^2-8xy\ge4+8-4-2.\left(x+y\right)^2=8-2.\left(x+y\right)^2\ge8-2=6\)

Dấu "=" xảy ra <=> \(x=y=\frac{1}{2}\)

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

\(\Rightarrow A\ge x\left(1-\frac{y}{2}\right)+y\left(1-\frac{z}{2}\right)+z\left(1-\frac{x}{2}\right)=\left(x+y+z\right)-\frac{xy+yz+zx}{2}\ge3-\frac{\frac{9}{3}}{2}=\frac{3}{2}\)

Dau '=' xay ra khi \(x=y=z=1\)

Vay \(A_{min}=\frac{3}{2}\)khi \(x=y=z=1\)

Bài này thì chắc cô si ngược dấu thôi:v

\(LHS=\Sigma\frac{x}{1+y^2}=\Sigma x.\left(1-\frac{y^2}{1+y^2}\right)\)

\(\ge\Sigma x\left(1-\frac{y}{2}\right)=x+y+z-\frac{xy+yz+zx}{2}\)

\(\ge x+y+z-\frac{\left(x+y+z\right)^2}{6}=\frac{3}{2}\)

P/s: check xem có ngược dấu chỗ nào ko:v

\(M^2=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{2xy}{\sqrt{yz}}+\frac{2yz}{\sqrt{zx}}+\frac{2xz}{\sqrt{yz}}=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{2x\sqrt{y}}{\sqrt{z}}+\frac{2y\sqrt{z}}{\sqrt{x}}+\frac{2z\sqrt{x}}{\sqrt{y}}\)

Áp dụng bđt Cô-si: \(\frac{x^2}{y}+\frac{x\sqrt{y}}{\sqrt{z}}+\frac{x\sqrt{y}}{\sqrt{z}}+z\ge4\sqrt[4]{\frac{x^2}{y}.\frac{x\sqrt{y}}{\sqrt{z}}.\frac{x\sqrt{y}}{\sqrt{z}}.z}=4x\)

tương tự \(\frac{y^2}{z}+\frac{y\sqrt{z}}{\sqrt{x}}+\frac{y\sqrt{z}}{\sqrt{x}}+x\ge4y\);\(\frac{z^2}{x}+\frac{z\sqrt{x}}{\sqrt{y}}+\frac{z\sqrt{x}}{\sqrt{y}}+y\ge4z\)

=>\(M^2+x+y+z\ge4\left(x+y+z\right)\Rightarrow M^2\ge3\left(x+y+z\right)\ge3.12=36\Rightarrow M\ge6\)

Dấu "=" xảy ra khi x=y=z=4

Vậy minM=6 khi x=y=z=4

b1: Áp dụng bđt Cauchy Schwarz dạng Engel ta được:

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

=>minP=1 <=> x=y=z=2/3

Cauchy ngược dấu:v

\(A\ge x\left(1-\frac{y}{2}\right)+y\left(1-\frac{z}{2}\right)+z\left(1-\frac{x}{2}\right)\)

\(=x+y+z-\frac{xy+yz+zx}{2}\ge3-\frac{\left(x+y+z\right)^2}{6}=\frac{3}{2}\)

Đẳng thức xảy ra khi x = y = z = 1.

P/s: Ko chắc~