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\(P=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{4xy}+4xy+\frac{5}{4xy}\)
\(P\ge\frac{4}{x^2+y^2+2xy}+2\sqrt{\frac{4xy}{4xy}}+\frac{5}{\left(x+y\right)^2}=4+2+5=11\)
\(P_{min}=11\) khi \(x=y=\frac{1}{2}\)
1/a/
\(A=\frac{2}{xy}+\frac{3}{x^2+y^2}=\left(\frac{1}{xy}+\frac{1}{xy}+\frac{4}{x^2+y^2}\right)-\frac{1}{x^2+y^2}\)
\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}-\frac{1}{\frac{\left(x+y\right)^2}{2}}=16-2=14\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
b/
\(4B=\frac{4}{x^2+y^2}+\frac{8}{xy}+16xy=\left(\frac{4}{x^2+y^2}+\frac{1}{xy}+\frac{1}{xy}\right)+\left(\frac{1}{xy}+16xy\right)+\frac{5}{xy}\)
\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}+2\sqrt{\frac{1}{xy}.16xy}+\frac{5}{\frac{\left(x+y\right)^2}{4}}\)
\(=16+8+20=44\)
\(\Rightarrow B\ge11\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
By Titu's Lemma we easy have:
\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)
\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)
\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)
\(=\frac{17}{4}\)
Mk xin b2 nha!
\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)
\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)
\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)
\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)
Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)
\(C=\frac{\left(x+y\right)^2-4xy}{xy}+\frac{4xy}{\left(x+y\right)^2}=\frac{\left(x+y\right)^2}{xy}+\frac{4xy}{\left(x+y\right)^2}-4\)
\(C=\frac{\left(x+y\right)^2}{4xy}+\frac{4xy}{\left(x+y\right)^2}+\frac{3\left(x+y\right)^2}{4xy}-4\)
\(C\ge2\sqrt{\frac{\left(x+y\right)^2.4xy}{4xy\left(x+y\right)^2}}+\frac{3.4xy}{4xy}-4=1\)
\(C_{min}=1\) khi \(x=y\)
Ta có : \(\frac{x}{4y^2+1}=x-\frac{4xy^2}{4y^2+1};\frac{y}{4x^2+1}=y-\frac{4x^2y}{4x^2+1}\)
Áp dụng BĐT Cauchy ta có :
\(4y^2+1\ge4y;4x^2+1\ge4x\)
\(\Rightarrow x-\frac{4xy^2}{4y^2+1}+y-\frac{4x^2y}{4x^2+1}\ge x-\frac{4xy^2}{4y}+y-\frac{4x^2y}{4x}\)
\(=x+y-2xy=2xy\)
Đến đây ta áp dụng BĐT phụ \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)
\(x+y=4xy\Leftrightarrow\frac{1}{xy}=\frac{4}{x+y}\le\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=4\)
\(\Leftrightarrow\frac{1}{xy}\le4\Leftrightarrow2xy\ge\frac{1}{2}\)
\(\Leftrightarrow\frac{x}{4y^2+1}+\frac{y}{4x^2+1}\ge\frac{1}{2}\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x=y\\4y^2=1\\4x^2=1\end{cases}\Leftrightarrow x=y=\frac{1}{2}}\)
Bạn trên đã chứng minh \(xy\ge\frac{1}{4}\) rồi nên mình xin phép không trình bày
Áp dụng BĐT Cauchy Schwarz ta dễ có:
\(LHS=\frac{x^2}{4xy^2+x}+\frac{y^2}{4x^2y+y}\)
\(\ge\frac{\left(x+y\right)^2}{4xy\left(x+y\right)+\left(x+y\right)}=\frac{\left(x+y\right)^2}{\left(x+y\right)^2+\left(x+y\right)}\)
Ta cần đi chứng minh:
\(\frac{\left(x+y\right)^2}{\left(x+y\right)^2+\left(x+y\right)}\ge\frac{1}{2}\)
\(\Leftrightarrow\left(x+y\right)^2\ge x+y\Leftrightarrow x+y\ge1\)
Điều này là hiển nhiên vì theo AM - GM ta có:\(x+y\ge2\sqrt{xy}=1\)
Vậy ta có đpcm
\(P=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\left(\frac{1}{4xy}+4xy\right)\)
\(\Rightarrow P\ge\frac{4}{x^2+y^2+2xy}+2\sqrt{\frac{4xy}{4xz}}=\frac{4}{1^2}+4=8\)
Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)