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1.\(N=x^2+\frac{1000}{x}+\frac{1000}{x}\ge3\sqrt[3]{\frac{x^2.1000.1000}{x^2}}\)
\(\Rightarrow N\ge300\)
Dấu "=" xảy ra \(\Leftrightarrow x^3=1000\Leftrightarrow x=10\)
2.\(P=\left(5x+\frac{12}{x}\right)+\left(3y+\frac{16}{y}\right)\ge2\sqrt{60}+2\sqrt{48}=4\sqrt{15}+8\sqrt{3}\)
Dấu "=" xảy ra \(\Leftrightarrow5x=\frac{12}{x};3y=\frac{16}{y}\Leftrightarrow x=\sqrt{\frac{12}{5}};y=\frac{4\sqrt{3}}{3}\)
\(\)
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\(Q=\frac{x^2+2x+17}{2\left(x+1\right)}=\frac{\left(x+1\right)^2+16}{2\left(x+1\right)}=\frac{x+1}{2}+\frac{8}{x+1}\ge2\sqrt{\frac{x+1}{2}.\frac{8}{x+1}}=4\)
Dấu "=" tại x = 3
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\(P\ge\frac{2}{\sqrt{xy}}\sqrt{1+x^2y^2}=2\sqrt{\frac{1+x^2y^2}{xy}}=2\sqrt{\frac{1}{xy}+xy}\)\(=2\sqrt{\frac{1}{16xy}+xy+\frac{15}{16xy}}\ge2\sqrt{\frac{1}{2}+\frac{15}{4\left(x+y\right)^2}}=\sqrt{17}.\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}.\)
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Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(P=\dfrac{1}{x}+\dfrac{2}{y}=\dfrac{1}{x}+\dfrac{4}{2y}=\dfrac{1^2}{x}+\dfrac{2^2}{2y}\)
\(\ge\dfrac{\left(1+2\right)^2}{x+2y}=\dfrac{3^2}{3}=3\)
Đẳng thức xảy ra khi \(x=y=1\)
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Áp dụng bđt Côsi:
\(B=\frac{3}{2}x^3+\frac{3}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}\ge5\sqrt[5]{\left(\frac{3}{2}x^3\right)^2.\left(\frac{1}{3x^2}\right)^3}=\frac{5}{\sqrt[5]{12}}\)
Dấu bằng xảy ra khi \(\frac{3}{2}x^3=\frac{1}{3x^2}\Leftrightarrow x^5=\frac{2}{9}\)\(\Leftrightarrow x=\sqrt[5]{\frac{2}{9}}\)
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\(1,A=\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\)
\(\ge\frac{4}{\left(x+y^2\right)}+\frac{1}{\frac{\left(x+y\right)^2}{2}}\ge\frac{4}{1}+\frac{2}{1}=6\)
Dấu "=" <=> x= y = 1/2
\(2,A=\frac{x^2+y^2}{xy}=\frac{x}{y}+\frac{y}{x}=\left(\frac{x}{9y}+\frac{y}{x}\right)+\frac{8x}{9y}\ge2\sqrt{\frac{x}{9y}.\frac{y}{x}}+\frac{8.3y}{9y}\)
\(=2\sqrt{\frac{1}{9}}+\frac{8.3}{9}=\frac{10}{3}\)
Dấu "=" <=> x = 3y
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điểm rơi xấu quá: x=\(\dfrac{\sqrt[3]{9}}{2}\); y=\(\sqrt[3]{9}\), z =\(2\sqrt[3]{9}\) (4x=2y=z)
\(N=\frac{x^2+2000}{x}=x+\frac{2000}{x}\ge2\sqrt{x.\frac{2000}{x}}=2\sqrt{2000}=40\sqrt{5}\)
Dấu "=" tại \(x=20\sqrt{5}\)