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Lời giải:
Ta có: $A=x^2+\frac{1}{y(x-y)}$. Đặt $x-y=a$ với $a>0$ thì áp dụng BĐT AM-GM ta có:
$A=(a+y)^2+\frac{1}{ay}\geq 4ay+\frac{1}{ay}\geq 2\sqrt{4ay.\frac{1}{ay}}=4$
Vậy $A_{\min}=4$ khi $x=\sqrt{2}; y=\frac{1}{\sqrt{2}}$
\(P=\dfrac{x}{\sqrt{x+y-x}}+\dfrac{y}{\sqrt{x+y-y}}=\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\)
\(=\dfrac{x^2}{x\sqrt{y}}+\dfrac{y^2}{y\sqrt{x}}\ge\dfrac{\left(x+y\right)^2}{x\sqrt{y}+y\sqrt{x}}=\dfrac{\left(x+y\right)^2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}\)
\(\ge\dfrac{\left(x+y\right)^2}{\dfrac{x+y}{2}.\left(1.\sqrt{x}+1.\sqrt{y}\right)}\ge\dfrac{\left(x+y\right)^2}{\dfrac{x+y}{2}.\sqrt{\left(1^2+1^2\right)\left(x+y\right)}}=\dfrac{1}{\dfrac{1}{2}\sqrt{2}}=\sqrt{2}\)
"=" khi x = y = 1/2
\(\Leftrightarrow2P=6x+4y+\dfrac{12}{x}+\dfrac{16}{y}\\ \Leftrightarrow2P=\left(\dfrac{12}{x}+3x\right)+\left(\dfrac{16}{y}+y\right)+3\left(x+y\right)\\ \Leftrightarrow2P\ge2\sqrt{36}+2\sqrt{16}+3\cdot6=12+8+18=38\\ \Leftrightarrow P\ge19\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}3x^2=12\\y^2=16\\x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
\(P=3\left(x+\dfrac{9}{x}\right)+\left(y+\dfrac{16}{y}\right)+\left(x+y\right)\)
\(P\ge3.2\sqrt{\dfrac{9x}{x}}+2\sqrt{\dfrac{16y}{y}}+7=33\)
\(P_{min}=33\) khi \(\left(x;y\right)=\left(3;4\right)\)
\(A=\dfrac{x^2+y^2}{xy}+\dfrac{xy}{x^2+y^2}=\dfrac{x^2+y^2}{4xy}+\dfrac{xy}{x^2+y^2}+\dfrac{3\left(x^2+y^2\right)}{4xy}\)
\(A\ge2\sqrt{\dfrac{\left(x^2+y^2\right)xy}{4xy\left(x^2+y^2\right)}}+\dfrac{3.2xy}{4xy}=\dfrac{5}{2}\)
Dấu "=" xảy ra khi \(x=y\)
\(C=\dfrac{\left(x+y\right)^2-4xy}{xy}+\dfrac{6xy}{\left(x+y\right)^2}=\dfrac{\left(x+y\right)^2}{xy}+\dfrac{6xy}{\left(x+y\right)^2}-4\)
\(C=\dfrac{3\left(x+y\right)^2}{8xy}+\dfrac{6xy}{\left(x+y\right)^2}+\dfrac{5\left(x+y\right)^2}{8xy}-4\)
\(C\ge2\sqrt{\dfrac{18xy\left(x+y\right)^2}{8xy\left(x+y\right)^2}}+\dfrac{5.4xy}{8xy}-4=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(x=y\)
Áp dụng bất đẳng thức AM - GM:
\(\dfrac{x^2}{y-1}+4\left(y-1\right)\ge2\sqrt{\dfrac{x^2}{y-1}.4\left(y-1\right)}\)
\(\Rightarrow\dfrac{x^2}{y-1}+4\left(y-1\right)\ge4x\).
Tương tự, \(\dfrac{y^2}{x-1}+4\left(x-1\right)\ge4y\).
Cộng vế với vế hai bđt trên rồi rút gọn ta được:
\(\dfrac{x^2}{y-1}+\dfrac{y^2}{x-1}\ge8\)
\(\Rightarrow P\ge8+2013=2021\).
Đẳng thức xảy ra khi x = y = 2.
Vậy....
ÁP dụng BĐT Mincopxki, ta có:
\(A\ge\sqrt{\left(x+y\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}\)
\(=\sqrt{\left(x+y\right)^2+\dfrac{\left(x+y\right)^2}{\left(xy\right)^2}}\)
\(\ge\sqrt{2\sqrt{\left(x+y\right)^2.\dfrac{\left(x+y\right)^2}{\left(xy\right)^2}}}=\sqrt{\dfrac{2\left(x+y\right)^2}{xy}}\) (cô si)
\(\ge\sqrt{\dfrac{2.4xy}{xy}}=\sqrt{8}=2\sqrt{2}\left(Côsi\right)\)
Min \(A=2\sqrt{2}\Leftrightarrow x=y\)
Ta có \(p=x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}=2\). Ta đi tìm GTNN của \(B=p+\dfrac{1}{p}\).
Do \(B=\dfrac{p}{4}+\dfrac{1}{p}+\dfrac{3p}{4}\) \(\ge2\sqrt{\dfrac{p}{4}.\dfrac{1}{p}}+\dfrac{3.2}{4}\) \(=\dfrac{5}{2}\). ĐTXR \(\Leftrightarrow\left\{{}\begin{matrix}x=y\\p=2\end{matrix}\right.\) \(\Leftrightarrow x=y=1\).
Vậy GTNN của B là \(\dfrac{5}{2}\) khi \(x=y=1\)
1)Ta có:
\(A=\left(x^2-4x+4\right)+x+\dfrac{4}{x}+2012=\left(x-2\right)^2+x+\dfrac{4}{x}+2012\)Theo bđt cô-si ta có:
\(x+\dfrac{4}{x}\ge2\sqrt{\dfrac{x.4}{x}}=4\)
\(\left(x-2\right)^2\ge0\)
\(\Rightarrow A\ge0+4+2012\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\x=\dfrac{4}{x}\end{matrix}\right.\Rightarrow x=2}\)
2)Ta có:
\(B=\left(y^2-4y+4\right)+3y+\dfrac{12}{y}+2012=\left(y-2\right)^2+3y+\dfrac{12}{y}+2012\)Áp dụng bđt cô si ta có:
\(3y+\dfrac{12}{y}\ge2\sqrt{\dfrac{3y.12}{y}}=12\)
\(\left(y-2\right)^2\ge0\)
\(\Rightarrow B\ge0+12+2012=2024\)
Dấu "=" xảy ra khi
\(\left\{{}\begin{matrix}\left(y-2\right)^2=0\\3y=\dfrac{12}{y}\end{matrix}\right.\Rightarrow y=2}\)