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\(P=\dfrac{1}{2\left(x^2+y^2\right)}+\dfrac{4}{xy}+2xy\)
\(\Leftrightarrow2P=\dfrac{1}{x^2+y^2}+\dfrac{8}{xy}+4xy\)
\(\Leftrightarrow2P=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{4xy}+4xy+\dfrac{29}{4xy}\)
Áp dụng BĐT AM - GM , ta có :
\(\Leftrightarrow\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{4xy}+4xy+\dfrac{29}{4xy}\ge\dfrac{2}{\sqrt{\left(x^2+y^2\right)2xy}}+2\sqrt{\dfrac{1}{4xy}.4xy}+\dfrac{29}{4xy}\)
\(\Leftrightarrow2P\ge\)\(\dfrac{2}{\sqrt{\left(x^2+y^2\right)2xy}}+2+\dfrac{29}{4xy}\ge\dfrac{4}{\left(x+y\right)^2}+2+\dfrac{29}{\left(x+y\right)^2}\)
\(\Leftrightarrow2P\ge2+4+29=35\)
\(\Leftrightarrow P\ge\dfrac{35}{2}\)
\(\Rightarrow P_{Min}=\dfrac{35}{2}\Leftrightarrow x=y=\dfrac{1}{2}\)
Vì \(\left\{{}\begin{matrix}x>y\\xy< 0\end{matrix}\right.\)\(\Rightarrow x>0>y\)
Đặt \(y=-z\left(z>0\right)\) thì ta có:
\(P=\left(x+z\right)^2+\left(x+z+\dfrac{1}{x}+\dfrac{1}{z}\right)^2\)
\(\ge\left(x+z\right)^2+\left(x+z+\dfrac{4}{x+z}\right)^2\)
Đặt \(x+z=a\) thì ta có:
\(P\ge a^2+\left(a+\dfrac{4}{a}\right)^2=2a^2+\dfrac{16}{a^2}+8\)
\(\ge8+2\sqrt{2a^2.\dfrac{16}{a^2}}=8+8\sqrt{2}\)
Dấu = xảy ra khi: \(\left\{{}\begin{matrix}x=z\\2a^2=\dfrac{16}{a^2}\end{matrix}\right.\)
\(\Rightarrow x=z=\dfrac{1}{\sqrt[4]{2}}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{\sqrt[4]{2}}\\y=-\dfrac{1}{\sqrt[4]{2}}\end{matrix}\right.\)
Lời giải:
Ta có:
\(P=\frac{1}{x}+\frac{2}{y}+\frac{3}{2x+y}\)
\(\Leftrightarrow P=\frac{2x+y}{xy}+\frac{3}{2x+y}=\frac{2x+y}{2}+\frac{3}{2x+y}\)
Áp dụng BĐT AM-GM:
\(2x+y\geq 2\sqrt{2xy}=2\sqrt{4}=4\)
Ta có:
\(P=\frac{2x+y}{2}+\frac{8}{2x+y}-\frac{5}{2x+y}\)
Áp dụng BĐT AM-GM: \(\frac{2x+y}{2}+\frac{8}{2x+y}\geq 2\sqrt{4}=4\) (1)
\(2x+y\geq 4\Rightarrow \frac{5}{2x+y}\leq \frac{5}{4}\Rightarrow -\frac{5}{2x+y}\geq \frac{-5}{4}\) (2)
Từ \((1);(2)\Rightarrow P\geq 4+\frac{-5}{4}=\frac{11}{4}\)
Vậy P min \(=\frac{11}{4}\Leftrightarrow (x,y)=(1,2 )\)
Bài 2. Áp dụng BĐT Cauchy dưới dạng Engel , ta có :
\(\dfrac{1}{x}+\dfrac{4}{y}+\dfrac{9}{z}\) ≥ \(\dfrac{\left(1+4+9\right)^2}{x+y+z}=196\)
⇒ \(P_{MIN}=196."="\) ⇔ \(x=y=z=\dfrac{1}{3}\)
cái này giống này - Here. Mỗi tội bài này Min=22 khi x=y=1/2
a) \(P=\dfrac{\left(x^2+2xy+9y^2\right)-\left(x+3y-2\sqrt{xy}\right)2\sqrt{xy}}{x+3y-2\sqrt{xy}}\)
\(=\dfrac{\left(x^2+6xy+9y^2\right)-\left(x+3y\right)2\sqrt{xy}}{x+3y-2\sqrt{xy}}\)
\(=\dfrac{\left(x+3y\right)^2-\left(x+3y\right)2\sqrt{xy}}{x+3y-2\sqrt{xy}}\)
\(=\dfrac{\left(x+3y\right)\left(x+3y-2\sqrt{xy}\right)}{x+3y-2\sqrt{xy}}\)
\(P=x+3y\)
b) \(\dfrac{P}{\sqrt{xy}+y}=\dfrac{x+3y}{\sqrt{xy}+y}=\dfrac{\left(x+3y\right):y}{\left(\sqrt{xy}+y\right):y}=\dfrac{\dfrac{x}{y}+3}{\sqrt{\dfrac{x}{y}}+1}\)
Đặt \(t=\sqrt{\dfrac{x}{y}}>0\) và \(\dfrac{P}{\sqrt{xy}+y}=Q\) thì \(Q=\dfrac{t^2+3}{t+1}=\dfrac{\left(t-1\right)^2+2\left(t+1\right)}{t+1}=2+\dfrac{\left(t-1\right)^2}{t+1}\ge2\)
\(Q_{min}=2\Leftrightarrow t=1\Leftrightarrow x=y\)
Lời giải:
Ta có: \(A=\frac{3}{x^2+y^2}+\frac{4}{xy}=3\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{5}{2xy}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{x^2+y^2}+\frac{1}{2xy}\geq \frac{4}{x^2+y^2+2xy}=\frac{4}{(x+y)^2}=4\)
Áp dụng BĐT Am-Gm: \(xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}\Rightarrow \frac{5}{2xy}\geq 10\)
Do đó: \(A\geq 3.4+10\Leftrightarrow A\geq 22\)
Vậy \(A_{\min}=22\Leftrightarrow x=y=\frac{1}{2}\)