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\(P=\sqrt{x^4+x^2y^2}+x^2=\sqrt{x^4+\frac{1}{x^2}}+x^2\)
Ta có: \(x^4+\frac{1}{x^2}=x^4+\frac{1}{8x^2}+\frac{1}{8x^2}+...+\frac{1}{8x^2}\ge9\sqrt[9]{x^4.\left(\frac{1}{8x^2}\right)^8}\)
\(=9\sqrt[9]{\frac{1}{8^8.x^{12}}}\)
=> \(P=3\sqrt[18]{\frac{1}{8^8.x^{12}}}+x^2\)
\(=\sqrt[18]{\frac{1}{8^8x^{12}}}+\sqrt[18]{\frac{1}{8^8x^{12}}}+\sqrt[18]{\frac{1}{8^8x^{12}}}+x^2\)
\(\ge4\sqrt[4]{\left(\sqrt[18]{\frac{1}{8^8x^{12}}}\right)^3.x^2}\)
\(=4.\left(\frac{1}{8^{\frac{1}{3}}.x^{\frac{1}{2}}}\right).x^2=2\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x^4=\frac{1}{8x^2}\\x^2=\sqrt[8]{\frac{1}{8^8x^{12}}}\end{cases}}\)<=> x^2 = 1/2 khi đó y = 2 , x = \(\frac{1}{\sqrt{2}}\)
Vậy GTNN của P = 2.
Bài 1:
\(\Leftrightarrow\left(x^2-6x-7\right)^2-\left(3x^2-12x-9\right)^2=0\)
\(\Leftrightarrow\left(3x^2-12x-9-x^2+6x+7\right)\left(3x^2-12x-9+x^2-6x-7\right)=0\)
\(\Leftrightarrow\left(2x^2-6x-2\right)\left(4x^2-18x-16\right)=0\)
\(\Leftrightarrow\left(x^2-3x-1\right)\left(2x^2-9x-8\right)=0\)
hay \(x\in\left\{\dfrac{3+\sqrt{13}}{2};\dfrac{3-\sqrt{13}}{2};\dfrac{9+\sqrt{145}}{4};\dfrac{9-\sqrt{145}}{4}\right\}\)
Bài này cũng dễ mà:
Áp dụng BĐT Cô-si, ta có:
\(y+z+1\ge3\sqrt[3]{yz}\)
\(\Rightarrow\)\(\dfrac{y+z+1}{3}\ge\sqrt[3]{yz}\)
\(\Rightarrow\)\(\dfrac{x}{\sqrt[3]{yz}}\ge\dfrac{3x}{y+z+1}\)
\(\Rightarrow\)\(\sum\dfrac{x}{\sqrt[3]{yz}}\ge\sum\dfrac{3x}{y+z+1}\)
Mà \(\sum\dfrac{3x}{y+z+1}=\sum\dfrac{3x^2}{xy+xz+x}\)
Áp dụng BĐT Cauchy -Schwaz:
\(\sum\dfrac{3x^2}{xy+xz+x}\ge\dfrac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\)
Mà:
\(xy+yz+xz\le x^2+y^2+z^2\)(BĐT phụ)
\(\Rightarrow\)\(2\left(xy+yz+xz\right)\le2\left(x^2+y^2+z^2\right)=6\)
Áp dụng BĐT Bunhicopski:
\(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\)
\(\Rightarrow x+y+z\le3\)
\(\Rightarrow2\left(xy+yz+xz\right)+x+y+z\le6+3=9\)
\(\Rightarrow\)\(\dfrac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{3\left(x+y+z\right)^2}{9}\ge\dfrac{\left(x+y+z\right)^2}{3}\ge xy+yz+xz\left(ĐPCM\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\)x=y=z=1
Sửa: \(P=2x^4+x^3\left(2y-1\right)+y^3\left(2x-1\right)+2y^4\); x+y=1
Ta có \(P=2x^4+x^3\left(2y-1\right)+y^3\left(2x-1\right)+2y^4=2x^4+2x^3y-x^3+2xy^3-y^3+2y^4\)
\(=x^3\left(2x+2y\right)+y^3\left(2x+2y\right)-\left(x^3+y^3\right)=\left(2x+2y\right)\left(x^3+y^3\right)-\left(x^3+y^3\right)\)
\(=\left(2x+2y-1\right)\left(x^3+y^3\right)=x^3+y^3\)
Do \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=x^2-xy+y^2=\frac{1}{2}\left(x^2+y^2\right)\left(\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}\right)^2\)
\(\Rightarrow P\ge\frac{1}{2}\left(x^2+y^2\right)\)
Mà \(x+y=1\Rightarrow x^2+y^2+2xy=1\Rightarrow2\left(x^2+y^2\right)-\left(x-y\right)^2=1\)
\(\Rightarrow2\left(x^2+y^2\right)\ge1\Rightarrow\left(x^2+y^2\right)\ge\frac{1}{2}\Rightarrow P\ge\frac{1}{4}\)
Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\ge\dfrac{x^3}{\dfrac{x^2+1-x^2}{2}}=2x^3\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{y^2}{\sqrt{1-y^2}}\ge2y^3;\dfrac{z^2}{\sqrt{1-z^2}}\ge2z^3\)
Cộng theo vế 3 BĐT trên ta có:
\(P\ge2x^3+2y^3+2z^3=2\left(x^3+y^3+z^3\right)=2\)
\(1=x+y+3xy\le x+y+\dfrac{3}{4}\left(x+y\right)^2\)
\(\Rightarrow3\left(x+y\right)^2+4\left(x+y\right)-4\ge0\)
\(\Rightarrow3\left(x+y+2\right)\left(x+y-\dfrac{2}{3}\right)\ge0\)
\(\Rightarrow x+y\ge\dfrac{2}{3}\) \(\Rightarrow\dfrac{1}{x+y}\le\dfrac{3}{2}\)
Đồng thời: \(x^2+y^2\ge\dfrac{1}{2}\left(x+y\right)^2\ge\dfrac{1}{2}.\left(\dfrac{2}{3}\right)^2=\dfrac{2}{9}\)
\(\Rightarrow-\left(x^2+y^2\right)\le-\dfrac{2}{9}\)
Từ đó ta có:
\(A=\sqrt{1-x^2}+\sqrt{1-y^2}+\dfrac{1-\left(x+y\right)}{x+y}=\sqrt{1-x^2}+\sqrt{1-y^2}+\dfrac{1}{x+y}-1\)
\(A\le\sqrt{2\left[2-\left(x^2+y^2\right)\right]}+\dfrac{1}{x+y}-1\le\sqrt{2\left(2-\dfrac{2}{9}\right)}+\dfrac{3}{2}-1=\dfrac{3+8\sqrt{2}}{6}\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{3}\)
\(\left(x^3+y^3\right)\left(x+y\right)=xy\left(1-x\right)\left(1-y\right)\Leftrightarrow\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)=\left(1-x\right)\left(1-y\right)\left(1\right)\)
Ta có : \(\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)\ge4xy\)
và \(\left(1-x\right)\left(1-y\right)=1-\left(x+y\right)+xy\le1-2\sqrt{xy}+xy\)
\(\Rightarrow1-2\sqrt{xy}+xy\ge4xy\Leftrightarrow0\) <\(xy\le\frac{1}{9}\)
Dễ chứng minh : \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\le\frac{1}{1+xy};\left(x,y\in\left(0;1\right)\right)\)
\(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\le\sqrt{2\left(\frac{1}{1+x^2}+\frac{1}{1+y^2}\right)}\le\sqrt{2\left(\frac{2}{1+xy}\right)}=\frac{2}{\sqrt{1+xy}}\)
\(3xy-\left(x^2+y^2\right)=xy-\left(x-y\right)^2\le xy\)
\(\Rightarrow P\le\frac{2}{\sqrt{1+xy}}+xy=\frac{2}{\sqrt{1+t}}+t\), \(\left(t=xy\right)\), (0<\(t\le\frac{1}{9}\)
Xét hàm số :
\(f\left(t\right)=\frac{2}{\sqrt{t+1}}+t\) , (0<\(t\le\frac{1}{9}\)
\(\sqrt{x+2017}-y^3=\sqrt{y+2017}-x^3\)
\(\Leftrightarrow\left(\sqrt{x+2017}-\sqrt{y+2017}\right)+\left(x^3-y^3\right)=0\)
\(\Leftrightarrow\dfrac{x-y}{\sqrt{x+2017}+\sqrt{y+2017}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(\dfrac{1}{\sqrt{x+2017}+\sqrt{y+2017}}+\left(x^2+xy+y^2\right)\right)=0\)
\(\Leftrightarrow x=y\)
\(\Rightarrow P=x^2-3x^2+12x-x^2+2018\)
\(=-3x^2+12x+2018=2030-3\left(x-2\right)^2\le2030\)