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a: \(x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
b: \(x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
c: \(=x^2-2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x,y\ne0\)
\(1>=\left(x+y\right)^2>=\left(2\sqrt{xy}\right)^2=4xy\Rightarrow1>=4xy\Rightarrow\frac{1}{2}>=2xy\)(bđt cosi)
\(\Rightarrow\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{2xy}>=\frac{4}{x^2+2xy+y^2}+\frac{1}{\frac{1}{2}}\)
\(=\frac{4}{\left(x+y\right)^2}+2>=\frac{4}{1^2}+2=4+2=6\)
dấu = xảy ra khi \(x=y=\frac{1}{2}\)
vậy min \(\frac{1}{x^2+y^2}+\frac{1}{xy}=6\)khi \(x=y=\frac{1}{2}\)
Áp dụng BDT AM-GM ta có:\(VT\ge3\left(\frac{x}{y+z+1}+\frac{y}{x+z+1}+\frac{z}{x+y+1}\right)\)
\(\Rightarrow\frac{VT}{3}\ge\frac{x^2}{xy+xz+x}+\frac{y^2}{yz+yx+y}+\frac{z^2}{xz+zy+z}\)
\(\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+xy+z}\) (Cauchy-Schwarz)
Do \(3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)\(\Rightarrow\left(x+y+z\right)^2\le\left(x^2+y^2+z^2\right)^2\)
\(\Rightarrow x+y+z\le x^2+y^2+z^2\).Suy ra
\(2\left(xy+yz+xz\right)+x+y+z\le2\left(xy+yz+xz\right)+x^2+y^2+z^2=\left(x+y+z\right)^2\)
Suy ra \(\frac{VT}{3}\le\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\Rightarrow VT\ge3\) (điều phải chứng minh)
Dấu "=" xảy ra khi x=y=z=1
Áp dụng BĐT AM-GM ta có:
\(\left\{{}\begin{matrix}1+x\ge2\sqrt{x}\\x+y\ge2\sqrt{xy}\\1+y\ge2\sqrt{y}\end{matrix}\right.\)
Cộng theo vế 3 BĐT trên ta có:
\(2\left(1+x+y\right)\ge2\left(\sqrt{x}+\sqrt{y}+\sqrt{xy}\right)\)
\(\Leftrightarrow VT=1+x+y\ge\sqrt{x}+\sqrt{y}+\sqrt{xy}=VP\)
Xảy ra khi \(\left\{{}\begin{matrix}1+x=2\sqrt{x}\\x+y=2\sqrt{xy}\\1+y=2\sqrt{y}\end{matrix}\right.\)\(\Rightarrow x=y=1\)
Khi đó \(P=x^2+y^2=1^2+1^2=2\)
Và \(Q=x^{2009}+y^{2009}=1^{2009}+1^{2009}=2\)
Với \(x,y>0\) ta có
\(1+x+y=\sqrt{x}+\sqrt{xy}+\sqrt{y}\)
\(\Leftrightarrow2+2x+2y-2\sqrt{x}-2\sqrt{xy}-2\sqrt{y}=0\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-2\sqrt{y}+1\right)+\left(x-2\sqrt{xy}+y\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2+\left(\sqrt{x}-\sqrt{y}\right)^2=0\)
\(\forall x,y>0\) ta luôn có \(\left\{{}\begin{matrix}\left(\sqrt{x}-1\right)^2\ge0\\\left(\sqrt{y}-1\right)^2\ge0\\\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2+\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\)
Đẳng thức xảy ra \(\Leftrightarrow x=y=1\)
Vậy x=y=1
Nên P=Q=2
1, A= y^3(1-y)^2 = 4/9 . y^3 . 9/4 (1-y)^2
= 4/9 .y.y.y . (3/2-3/2.y)^2
=4/9 .y.y.y (3/2-3/2.y)(3/2-3/2.y)
<= 4/9 (y+y+y+3/2-3/2.y+3/2-3/2.y)^5
=4/9 . 243/3125
=108/3125
Đến đó tự giải
2.
\(x+y+1=\sqrt{x}+\sqrt{y}+\sqrt{xy}\)
\(\Leftrightarrow2x+2y+2=2\sqrt{x}+2\sqrt{y}+2\sqrt{xy}\)
\(\Leftrightarrow x-2\sqrt{xy}+y+x-2\sqrt{x}+1+y-2\sqrt{y}+1=0\)
\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=\sqrt{y}\\\sqrt{x}=1\\\sqrt{y}=1\end{matrix}\right.\Leftrightarrow x=y=1\)
Từ đó suy ra : \(\left\{{}\begin{matrix}P=1^2+1^2=2\\Q=1^{1023}+1^{2014}=2\end{matrix}\right.\)
1.
Xét \(x^3+y^3+xy=\left(x+y\right)\left(x^2-xy+y^2\right)+xy\)
\(=x^2-xy+y^2+xy\)( vì \(x+y=1\))
\(=x^2+y^2\)
Áp dụng bất đẳng thức Bunhiacopxki :
\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x+y\right)^2=1\)
\(\Rightarrow x^2+y^2\ge\frac{1}{2}\)
Từ đó ta có : \(P=\frac{1}{x^2+y^2}\le\frac{1}{\frac{1}{2}}=2\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)
\(\left(\dfrac{x^2+y^2}{x-y}\right)^2=\dfrac{\left(x^2+y^2\right)^2}{x^2+y^2-2}\)
Đặt \(x^2+y^2=a\left(a>0\right)\)ta sẽ chứng minh \(\dfrac{a^2}{a-2}\ge8\Leftrightarrow\left(a-4\right)^2\ge0\)(đúng)
\(\Rightarrow\dfrac{x^2+y^2}{x-y}\ge2\sqrt{2}\)
Dấu = xảy ra khi \(\left\{{}\begin{matrix}xy=1\\x^2+y^2=4\end{matrix}\right.\)( giải ra )