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\(1=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le\sqrt{3}\)
\(P=\sum\frac{1}{\sqrt{\left(2x+y\right)^2+\left(x-y\right)^2}}\le\sum\frac{1}{\sqrt{\left(2x+y\right)^2}}=\sum\frac{1}{2x+y}\)
\(P\le\sum\left(\frac{1}{x+x+y}\right)\le\frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\le\frac{\sqrt{3}}{3}\)
\(\Rightarrow P_{max}=\frac{\sqrt{3}}{3}\) khi \(x=y=z=\sqrt{3}\)
ta có: \(\frac{\sqrt{2x^2+y^2}}{xy}=\sqrt{\frac{2}{y^2}+\frac{1}{x^2}}\)
Áp dụng BĐT bunyakovsky:\(\left(2+1\right)\left(\frac{2}{y^2}+\frac{1}{x^2}\right)\ge\left(\frac{2}{y}+\frac{1}{x}\right)^2\)
\(\Rightarrow\frac{2}{y^2}+\frac{1}{x^2}\ge\frac{1}{3}\left(\frac{2}{y}+\frac{1}{x}\right)^2\).....bla bla
Ta có:
\(15\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)=10\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)+2014\)
\(\le10\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2014\)
=> \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\le\frac{2014}{5}\)
\(P=\frac{1}{\sqrt{5x^2+2xy+2yz}}+\frac{1}{\sqrt{5y^2+2yz+2zx}}+\frac{1}{\sqrt{5z^2+2zx+2xy}}\)
=> \(P\sqrt{\frac{2014}{135}}=\frac{1}{\sqrt{5x^2+2xy+2yz}.\sqrt{\frac{135}{2014}}}\)
\(+\frac{1}{\sqrt{5y^2+2yz+2zx}\sqrt{\frac{135}{2014}}}+\frac{1}{\sqrt{\frac{135}{2014}}\sqrt{5z^2+2zx+2xy}}\)
\(\le\frac{1}{2}\left(\frac{1}{5x^2+2xy+2yz}+\frac{2014}{135}+\frac{1}{5y^2+2yz+2zx}+\frac{2024}{135}+\frac{1}{5z^2+2yz+2zx}+\frac{2014}{135}\right)\)
\(\le\frac{1}{2}\left[\frac{1}{81}\left(\frac{5}{x^2}+\frac{2}{xy}+\frac{2}{yz}\right)+\frac{1}{81}\left(\frac{5}{y^2}+\frac{2}{yz}+\frac{2}{zx}\right)+\frac{1}{81}\left(\frac{5}{z^2}+\frac{2}{zx}+\frac{2}{xy}\right)+\frac{2014}{45}\right]\)
\(=\frac{5}{162}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+\frac{2}{81}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)+\frac{1007}{45}\)
\(\le\frac{5}{162}.\frac{2014}{5}+\frac{2}{81}.\frac{2014}{5}+\frac{1007}{45}=\frac{2014}{45}\)
=> \(P\le\frac{2014}{45}:\sqrt{\frac{2014}{135}}=3\sqrt{\frac{2014}{135}}\)
Dấu "=" xảy ra <=> x = y = z = \(\sqrt{\frac{15}{2014}}\)
1) đặt \(\sqrt{x-1}=a\left(a\ge0\right);\sqrt{y-4}=b\left(b\ge0;\right)\)
M = \(\frac{a}{a^2+1}+\frac{b}{b^2+4}\); a2 +1 \(\ge2a;b^2+4\ge4b\)=> M \(\le\frac{a}{2a}+\frac{b}{4b}=\frac{3}{4}\)
M đạt GTLN khi a=1, b=2 hay x=2; y= 8
2) <=> (x-y)2 + (x+2)2 =8 => (x+2)2\(\le8< =>\left|x+2\right|\le\sqrt{8}\approx2< =>-2\le x+2\le2< =>\)\(-4\le x\le0\)
x=-4 => (y+4)2 =4 <=> y = -2;y = -6
x=-3 => (y+3)2 = 7 (vô nghiệm); x=-1 => (y+1)2 =7 (vô nghiệm)
x=0 => y2 = 4 => y =2; =-2
vậy có các nghiệm (x;y) = (-4;-2); (-4;-6); (0;-2); (0;2)
3) \(\frac{x^2}{y^2}+\frac{y^2}{z^2}\ge2\frac{x}{z}\left(a^2+b^2\ge2ab\right)\); tương tự với các số còn lại ta được điều phải chứng minh
3) sửa lại
áp dụng a2+b2+c2 \(\ge\frac{\left(a+b+c\right)^2}{3}\)
\(\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}\ge\frac{\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)^2}{3}\ge\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\)(vì \(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge3\sqrt[3]{\frac{xyz}{yzx}}=3\))
dấu '=' khi x=y=z
Ta có 5x2+2xy+2y2=(2x+y)2+(x-y)2>=(2x+y)2
Khi đó P<=\(\frac{1}{2x+y}+\frac{1}{2y+z}+\frac{1}{2z+x}\)
Lại có \(\frac{1}{2x+y}=\frac{1}{x+x+y}\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}\right)\)
Tương tự \(\frac{1}{2y+z}\le\frac{1}{9}\left(\frac{1}{y}+\frac{1}{z}+\frac{1}{y}\right)\)
\(\frac{1}{2z+x}\le\frac{1}{9}\left(\frac{1}{z}+\frac{1}{x}+\frac{1}{z}\right)\)
Khi đó P<=\(\frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\le\frac{1}{3}\sqrt{3\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)}\le\frac{\sqrt{3}}{3}\)
Dấu bằng xảy ra khi x=y=z=\(\frac{\sqrt{3}}{3}\)
HAY
bài làm láo à ? sau 1 hồi trình bày thì dấu = khi \(x=y=z=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\) ??