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Ta có: \(x^2+\frac{1}{4}\ge x\Rightarrow x^2+y+\frac{3}{4}\ge x+y+\frac{1}{2}\)
Tương tự \(y^2+x+\frac{3}{4}\ge x+y+\frac{1}{2}\)
\(\Rightarrow\left(x^2+y+\frac{3}{4}\right)\left(y^2+x+\frac{3}{4}\right)\ge\left(x+y+\frac{1}{2}\right)^2\) (1)
Mặt khác: \(\left(2x+\frac{1}{2}\right)\left(2y+\frac{1}{2}\right)\le\frac{1}{4}\left(2x+2y+1\right)^2=\left(x+y+\frac{1}{2}\right)^2\) (2)
(1);(2) \(\Rightarrow\left(x^2+y+\frac{3}{4}\right)\left(y^2+x+\frac{3}{4}\right)\ge\left(2x+\frac{1}{2}\right)\left(2y+\frac{1}{2}\right)\)
Dấu "=" xảy ra khi và chỉ khi \(x=y=\frac{1}{2}\)
\(x^3+3x^2+3x+1+y^3+3y^3+3y+1+x+y+2=0\)
\(\Leftrightarrow\left(x+1\right)^3+\left(y+1\right)^3+x+y+2=0\)
\(\Leftrightarrow\left(x+y+2\right)\left(\left(x+1\right)^2+\left(y+1\right)^2-\left(x+1\right)\left(y+1\right)\right)+\left(x+y+2\right)=0\)
\(\Leftrightarrow\left(x+y+2\right)\left(\left(x+1\right)^2+\left(y+1\right)^2-\left(x+1\right)\left(y+1\right)+1\right)=0\)
\(\Leftrightarrow x+y+2=0\)
(phần trong ngoặc \(\left(x+1\right)^2-\left(x+1\right)\left(y+1\right)+\frac{\left(y+1\right)^2}{4}+\frac{3\left(y+1\right)^2}{4}+1\)
\(=\left(x+1-\frac{y+1}{4}\right)^2+\frac{3\left(y+1\right)^2}{4}+1\) luôn dương)
\(\Rightarrow x+y=-2\)
Mà \(xy>0\Rightarrow\left\{{}\begin{matrix}x< 0\\y< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-x>0\\-y>0\end{matrix}\right.\)
Ta có: \(\frac{1}{-x}+\frac{1}{-y}\ge\frac{4}{-\left(x+y\right)}=2\) \(\Leftrightarrow\frac{1}{x}+\frac{1}{y}\le-2\) (đpcm)
Dấu "=" xảy ra khi và chỉ khi \(x=y=-1\)
2/ \(x;y;z\ne0\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{1}{z}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{xz+yz+z^2}=0\)
\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{xz+yz+z^2}\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(\frac{xy+yz+xz+z^2}{xyz\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\) dù trường hợp nào thì thay vào ta đều có \(B=0\)
3/ \(\Leftrightarrow mx-2x+my-y-1=0\)
\(\Leftrightarrow m\left(x+y\right)-\left(2x+y+1\right)=0\)
Gọi \(A\left(x_0;y_0\right)\) là điểm cố định mà d đi qua
\(\Leftrightarrow\left\{{}\begin{matrix}x_0+y_0=0\\2x_0+y_0+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_0=-1\\y_0=1\end{matrix}\right.\)
Vậy d luôn đi qua \(A\left(-1;1\right)\) với mọi m
Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\Leftrightarrow\left(x+y\right)\left(\frac{zx+z^2+zy+xy}{xyz\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left[z\left(x+z\right)+y\left(x+z\right)\right]=0\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Rightarrow\left(x^2-y^2\right)\left(y^3+z^3\right)\left(z^4-x^4\right)=0\).
Vậy \(M=\frac{3}{4}+\left(x^2-y^2\right)\left(y^3+z^3\right)\left(z^4-x^4\right)=\frac{3}{4}+0=\frac{3}{4}\)
Mới nghĩ ra 3 câu:
a/ \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}=\frac{ab}{\sqrt{\left(a+b\right)^2\left(1+c\right)}}\le\frac{ab}{2\sqrt{ab\left(1+c\right)}}=\frac{1}{2}\sqrt{\frac{ab}{1+c}}\)
\(\sum\sqrt{\frac{ab}{1+c}}\le\sqrt{2\sum\frac{ab}{1+c}}\)
\(\sum\frac{ab}{1+c}=\sum\frac{ab}{a+c+b+c}\le\frac{1}{4}\sum\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)=\frac{1}{4}\)
c/ \(ab+bc+ca=2abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\Rightarrow x+y+z=2\)
\(VT=\sum\frac{x^3}{\left(2-x\right)^2}\)
Ta có đánh giá: \(\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\) \(\forall x\in\left(0;2\right)\)
\(\Leftrightarrow2x^3\ge\left(2x-1\right)\left(x^2-4x+4\right)\)
\(\Leftrightarrow9x^2-12x+4\ge0\Leftrightarrow\left(3x-2\right)^2\ge0\)
d/ Ta có đánh giá: \(\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\)
\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)
Akai Haruma, Nguyễn Ngọc Lộc , @tth_new, @Băng Băng 2k6, @Trần Thanh Phương, @Nguyễn Việt Lâm
Mn giúp e vs ạ! Thanks!
