\(VT=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\left(\frac{1}{4xy}+4xy\right)+\frac{5}{4xy}\)

\(\ge\frac{4}{\left(x+y\right)^2}+2+\frac{5}{\left(x+y\right)^2}\ge4+2+5=11\)

Với điều kiện \(ab+bc+ca+abc=4\) thì \(VP-VT=\frac{bc^2\left(a-b\right)^2+ca^2\left(b-c\right)^2+ab^2\left(c-a\right)^2}{\left(a^2+2b\right)\left(b^2+2c\right)\left(c^2+2a\right)}\ge0\)

Cauchy ngược dấu + Svacxo + gt coi 

moi hok lop 6

may dua con nit bien ra cho khac choi

hình như đề sai

(2x + y)x \(\le2x\) <=> \(2x^2+xy\le2x\)(1)

Vì \(0\le x\le y\Leftrightarrow y-x\ge0\) mà \(y\le1\Rightarrow\left(y-x\right)y\le y-x\) (2)

Lấy (1) + (2) => \(2x^2+y^2\le x+y\)

áp dụng BĐT bun nhi a cốp xki :

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

Vì \(2x^2+y^2\ge0\) chia cả hai vế cho 2x^ 2 + y^2 ta đc ĐPCM . Dấu = xảy ra khi .... ( tự tìm )

ko giai di ra cho khac

nhiều trường hợp xảy ra dấu bằng thì sao

\(\frac{3}{2}x^2+y^2+z^2+yz=1\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)

\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)=2\)

\(\Leftrightarrow\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2=2\)

Suy ra : \(A^2\le2\Rightarrow A\le\sqrt{2}\)

Vậy Max A = \(\sqrt{2}\) khi \(\hept{\begin{cases}x=y\\x=z\\x+y+z=\sqrt{2}\end{cases}\Leftrightarrow}x=y=z=\frac{\sqrt{2}}{3}\)

tuyệt

srweafgtseawref

\(Gt\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\Rightarrow ab+bc+ca=1\)

\(VT=\frac{2}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}+\frac{1}{\sqrt{1+z^2}}\)

\(=\frac{\frac{2}{x}}{\sqrt{\frac{1}{x^2}+1}}+\frac{\frac{1}{y}}{\sqrt{\frac{1}{y^2}+1}}+\frac{\frac{1}{z}}{\sqrt{\frac{1}{z^2}+1}}\)

\(=\frac{2a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+ab+bc+ca}}+\frac{c}{\sqrt{c^2+ab+bc+ca}}\)

\(=\sqrt{\frac{2a}{\left(a+b\right)}\cdot\frac{2a}{\left(a+c\right)}}+\sqrt{\frac{2b}{\left(b+a\right)}\cdot\frac{b}{2\left(b+c\right)}}\)\(+\sqrt{\frac{2c}{\left(c+a\right)}\cdot\frac{c}{2\left(c+b\right)}}\)

\(\le\frac{\frac{2a}{a+b}+\frac{2a}{a+c}+\frac{2b}{a+b}+\frac{b}{2\left(b+c\right)}+\frac{2c}{c+a}+\frac{c}{2\left(c+b\right)}}{2}=\frac{9}{4}\)

\(x^3+y^3+y^3\ge3\sqrt[3]{x^3.y^3.y^3}=3xy^2\)

\(x^3+1+1\ge3x\)

\(2\left(y^3+1+1\right)\ge6y\)

Cộng vế:

\(2\left(x^3+2y^3\right)+6\ge3\left(x+2y+xy^2\right)=12\)

\(\Rightarrow x^3+2y^3\ge3\) (đpcm)

Dấu "=" xảy ra khi \(x=y=1\)