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Ta có: \(\frac{1}{x}+\frac{2}{y}=2\ge2\sqrt{\frac{2}{xy}}\Leftrightarrow\sqrt{\frac{2}{xy}}\le1\Leftrightarrow xy\ge2\)
\(5x^2+y-4xy+y^2=\left(2x-y\right)^2+x^2+y\)
\(\ge x^2+y=x^2+\frac{y}{2}+\frac{y}{2}\ge3\sqrt[3]{\frac{\left(xy\right)^2}{4}}\ge3\left(đpcm\right)\)
Dấu "="\(\Leftrightarrow x=1,y=2\)
Ta có:
\(\frac{1}{x}+\frac{2}{y}=2\ge2\sqrt{\frac{2}{xy}}\Rightarrow\sqrt{\frac{2}{xy}}\le1\Rightarrow xy\ge2\)
\(5x^2+y-4xy+y^2=\left(2x-y\right)^2+x^2+y\)
\(\ge x^2+y=x^2+\frac{y}{2}+\frac{y}{2}\)\(\ge3\sqrt[3]{\frac{\left(xy\right)^2}{4}}\ge3\)(Đpcm0
Dấu = khi x=1;y=2
Ta có:
\(\frac{1}{x}+\frac{2}{y}=2\ge2\sqrt{\frac{2}{xy}}\Rightarrow\sqrt{\frac{2}{xy}}\le1\Rightarrow xy\ge2\)
\(5x^2+y-4xy+y^2=\left(2x-y\right)^2+x^2+y\)
\(\ge x^2+y=x^2+\frac{y}{2}+\frac{y}{2}\ge3\sqrt[3]{\frac{\left(xy\right)^2}{4}}\ge3\)(Đpcm)
Dấu = khi x=1;y=2
Áp dụng BĐT Cauchy cho 2 số không âm, ta được:
\(\frac{1}{x}+\frac{2}{y}=2\ge2\sqrt{\frac{2}{xy}}\Leftrightarrow\sqrt{\frac{2}{xy}}\le1\Leftrightarrow xy\ge2\)
\(5x^2+y-4xy+y^2=\left(2x-y\right)^2+x^2+y\ge x^2+y\)
\(=x^2+\frac{y}{2}+\frac{y}{2}\ge3\sqrt[3]{x^2.\frac{y}{2}.\frac{y}{2}}=3\sqrt[3]{\frac{\left(xy\right)^2}{4}}\ge3\sqrt[3]{\frac{4}{4}}=3.1=3\)
Áp dụng BĐT Cauchy và Cauchy - Schwarz ta có:
\(\frac{1}{x^2+y^2}+\frac{2}{xy}+4xy\)
\(=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\left(4xy+\frac{1}{4xy}\right)+\frac{5}{4xy}\)
\(\ge\frac{4}{x^2+y^2+2xy}+2\sqrt{4xy\cdot\frac{1}{4xy}}+\frac{5}{\left(x+y\right)^2}\)
\(=\frac{4}{\left(x+y\right)^2}+2+\frac{5}{1^2}=4+2+5=11\)
Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)
Xét \(P=x^2+y^2+2x\left(y-1\right)+2y+1\)
\(P=x^2+y^2+2xy-2x+2y+1\)
+) Nếu \(y>x\) thì \(2y-2x+1>0\). Do đó \(P>\left(x+y\right)^2\). Hơn nữa:
\(P< x^2+y^2+1+2xy+2x+2y\) \(=\left(x+y+1\right)^2\),
suy ra \(\left(x+y\right)^2< P< \left(x+y+1\right)^2\), vô lí vì P là SCP.
+) Nếu \(x>y\) thì \(2y-2x+1< 0\) nên \(P< \left(x+y\right)^2\)
Hơn nữa \(P>x^2+y^2+1+2xy-2x-2y\) \(=\left(x+y-1\right)^2\)
Suy ra \(\left(x+y-1\right)^2< P< \left(x+y\right)^2\), vô lí vì P là SCP.
Vậy \(x=y\) (đpcm)
(Cơ mà nếu thay \(x=y\) vào P thì \(P=4x^2+1\) lại không phải là SCP đâu)
Áp dụng BĐT Cauchy cho 3 số dương, ta được:
\(\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\ge\sqrt[3]{\frac{1}{x\left(x+1\right)}.\frac{x}{2}.\frac{x+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)
\(\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{4}\ge\sqrt[3]{\frac{1}{y\left(y+1\right)}.\frac{y}{2}.\frac{y+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)
\(\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge\sqrt[3]{\frac{1}{z\left(z+1\right)}.\frac{z}{2}.\frac{z+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)
\(\Rightarrow\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\)\(+\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{4}\)
\(+\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge\frac{3}{2}.3=\frac{9}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{x+y+z}{2}+\frac{x+y+z+3}{4}\ge\frac{9}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{3}{2}+\frac{3}{2}\ge\frac{9}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}\ge\frac{3}{2}\left(đpcm\right)\)