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

Ta có:

\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge\frac{3a}{4}\)

\(\Leftrightarrow\frac{a^3}{\left(1+b\right)\left(1+c\right)}\ge\frac{6a-b-c-2}{8}\)

Tương tự ta có: \(\hept{\begin{cases}\frac{b^3}{\left(1+c\right)\left(1+a\right)}\ge\frac{6b-c-a-2}{8}\\\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{6c-a-b-2}{8}\end{cases}}\)

Cộng vế theo vế ta được

\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{6a-b-c-2}{8}+\frac{6b-c-a-2}{8}+\frac{6c-a-b-2}{8}\)

\(=\frac{a+b+c}{2}-\frac{3}{4}\ge\frac{3}{2}.\sqrt[3]{abc}-\frac{3}{4}=\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\)

Mai mình làm cho

Đặt biểu thức trên là A

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

\(=\left(x-y\right)^2+\frac{\left(xy-1\right)^2}{\left(x-y\right)^2}+2xy\ge2\sqrt{\left(x-y\right)^2\frac{\left(xy-1\right)^2}{\left(x-y\right)^2}}+2xy\)

\(=2\sqrt{\left(xy-1\right)^2}+2xy\)

\(=2\left|xy-1\right|+2xy\)

Áp dụng bđt Cô si 

- Nếu thấy \(xy\ge1\Rightarrow A\ge2xy-2+2xy=4xy-2\ge2\)

- Nếu \(xy< 1\Rightarrow A>-2xy+2+2xy=2\)

Vậy : \(A\ge2\left(đpcm\right)\)

Ta có:Xét hiệu \(x^2+y^2+\left(\frac{xy-1}{x-y}\right)^2-2=\left(x-y\right)^2+\left(\frac{xy-1}{x-y}\right)^2+2\left(xy-1\right)\ge0\)

\(=\left(x-y+\frac{xy-1}{x-y}\right)^2\ge0\)

\(\Rightarrow x^2+y^2+\left(\frac{xy-1}{x-y}\right)^2\ge2\left(đpcm\right)\)

Bài 1: Áp dụng BĐT AM-GM ta có:

\(1+x\ge2\sqrt{x}\)

\(x+y\ge2\sqrt{xy}\)

\(y+1\ge2\sqrt{y}\)

Cộng theo vế 3 BĐT trên ta có:

\(2\left(1+x+y\right)\ge2\left(\sqrt{x}+\sqrt{xy}+\sqrt{y}\right)\)

\(1+x+y\ge\sqrt{x}+\sqrt{xy}+\sqrt{y}\Leftrightarrow VT\ge VP\) 

Đẳng thức xảy ra khi  \(\hept{\begin{cases}1+x=2\sqrt{x}\\x+y=2\sqrt{xy}\\y+1=2\sqrt{y}\end{cases}}\Rightarrow x=y=1\)

Khi đó \(S=x^{2013}+y^{2013}=1^{2013}+1^{2013}=2\)

Bài 2: Vì \(\hept{\begin{cases}x,y,z\in\left[-1;3\right]\\x+y+z=3\end{cases}}\) nên 

\(0\le\left(x+1\right)\left(y+1\right)\left(z+1\right)+\left(3-x\right)\left(3-y\right)\left(3-z\right)\)

\(\Leftrightarrow0\le4\left(xy+yz+xz\right)-8\left(x+y+z\right)+28\)

\(\Leftrightarrow0\le2\left(xy+yz+xz\right)+2\)

\(\Leftrightarrow x^2+y^2+z^2\le x^2+y^2+z^2+2\left(xy+yz+xz\right)+2\)

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

\(\Leftrightarrow x^2+y^2+z^2\le3^2+2=9+2=11\)

Cảm ơn b Thắng Nguyễn

a/ Sửa đề:

\(\sqrt{22x^2+36xy+6y^2}+\sqrt{22y^2+36xy+6x^2}=x^2+y^2+32\)

\(\Leftrightarrow64x^2+64y^2+2048-64\sqrt{22x^2+36xy+6y^2}-64\sqrt{22y^2+36xy+6x^2}=0\)

\(\Leftrightarrow\left(22x^2+36xy+6y^2-64\sqrt{22x^2+36xy+6y^2}+1024\right)+\left(22y^2+36xy+6x^2-64\sqrt{22y^2+36xy+6x^2}+1024\right)+\left(36x^2-72xy+36y^2\right)=0\)

\(\Leftrightarrow\left(\sqrt{22x^2+36xy+y^2}-32\right)^2+\left(\sqrt{22y^2+36xy+6x^2}-32\right)^2+36\left(x-y\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{22x^2+36xy+6y^2}=32\\\sqrt{22y^2+36xy+6x^2}=32\\x=y\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{64x^2}=32\\x=y\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=y=4\\x=y=-4\end{cases}}\)

Câu b đề sai rồi.

Ta có:

\(\left(y^2+y+1\right)\left(x^2+x+1\right)\)

\(=x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+x+y+1\)

\(=x^2y^2+x^2+y^2+2xy+2=x^2y^2+3\)

Ta lại có:

\(\left(y^2+y+1\right)-\left(x^2+x+1\right)=\left(y^2-x^2\right)+\left(y-x\right)\)

\(=\left(y-x\right)\left(x+y+1\right)=-2\left(x-y\right)\)

Theo đề bài ta có: (sửa đề luôn)

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

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

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

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

\(=-\frac{2\left(x-y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)

kết bạn với mình nhé!

ai biết làm giúp với

\(A=\dfrac{2\left(x^3+y^3\right)}{\left(x^4+y^2\right)\left(x^2+y^4\right)}=2.\dfrac{\left(x^3+y^3\right)}{x^4y^4+x^2y^2+x^6+y^6}\)

\(=2.\dfrac{\left(x^3+y^3\right)}{1+1+x^6+y^6}=2.\dfrac{x^3+y^3}{x^6+y^6+2x^3y^3}=2.\dfrac{x^3+y^3}{\left(x^3+y^3\right)^2}=\dfrac{2}{x^3+y^3}\left(1\right)\)

Áp dụng bất đẳng thức Cauchy ta có:

\(x^3+y^3+1\ge3\sqrt{xy.1}=3\)

\(\Rightarrow x^3+y^3\ge2\Rightarrow\dfrac{2}{x^3+y^3}\le1\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow A\le1\)

Dấu "=" xảy ra khi x=y=1.

Vậy MaxA là 1, đạt được khi x=y=1.

Thanks!

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