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\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}=\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\dfrac{1}{2xy}\)
Áp dụng BĐT Schwarz : \(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\ge\dfrac{\left(1+1\right)^2}{x^2+y^2+2xy}=\dfrac{4}{\left(x+y\right)^2}=4\)
Lại có \(\dfrac{1}{2xy}=\dfrac{2}{4xy}\ge\dfrac{2}{\left(x+y\right)^2}=2\)
Cộng vế với vế được P \(\ge6\) ("=" khi x = y = 1/2)
Vậy Min P = 6 <=> x = y = 1/2
\(P=\dfrac{y}{x}+\dfrac{x}{y}+\left(\dfrac{x}{3y}+3xy+\dfrac{1}{3}+\dfrac{1}{3}\right)+12\left(xy+\dfrac{1}{9}\right)-2\)
\(P\ge2\sqrt{\dfrac{xy}{xy}}+4\sqrt[4]{\dfrac{3x^2y}{27y}}+12.2\sqrt{\dfrac{xy}{9}}-2\)
\(P\ge4\sqrt{\dfrac{x}{3}}+8\sqrt{xy}=4\left(2\sqrt{xy}+\sqrt{\dfrac{x}{3}}\right)=4\)
\(P_{min}=4\) khi \(x=y=\dfrac{1}{3}\)
\(\dfrac{\left(x+y+1\right)^2}{xy+x+y}\ge\dfrac{3\left(xy+x+y\right)}{xy+x+y}=3\)
\(\Rightarrow A=\dfrac{8\left(x+y+1\right)^2}{9\left(xy+x+y\right)}+\dfrac{\left(x+y+1\right)^2}{9\left(xy+x+y\right)}+\dfrac{xy+x+y}{\left(x+y+1\right)^2}\)
\(A\ge\dfrac{8}{9}.3+2\sqrt{\dfrac{\left(x+y+1\right)^2\left(xy+x+y\right)}{\left(xy+x+y\right)\left(x+y+1\right)^2}}=\dfrac{10}{3}\)
Dấu "=" xảy ra khi \(x=y=1\)
=t (t>=3). cho dễ làm
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: \(\left\{{}\begin{matrix}x=z\\2a^2=\dfrac{16}{a^2}\end{matrix}\right.\)
\(\Rightarrow x=z=\dfrac{1}{\sqrt[4]{2}}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{\sqrt[4]{2}}\\y=-\dfrac{1}{\sqrt[4]{2}}\end{matrix}\right.\)
\(A\ge\dfrac{\left(x+y\right)^2}{2xy}+\dfrac{\sqrt{xy}}{x+y}\)
\(A\ge\dfrac{7\left(x+y\right)^2}{16xy}+\dfrac{\left(x+y\right)^2}{16xy}+\dfrac{\sqrt{xy}}{2\left(x+y\right)}+\dfrac{\sqrt{xy}}{2\left(x+y\right)}\)
\(A\ge\dfrac{7.4xy}{16xy}+3\sqrt[3]{\dfrac{\left(x+y\right)^2xy}{16.4.xy\left(x+y\right)^2}}=\dfrac{5}{2}\)
Dấu "=" xảy ra khi \(x=y\)
\(\dfrac{x^3}{4\left(y+2\right)}+\dfrac{x\left(y+2\right)}{16}\ge\dfrac{x^2}{4}\) ; \(\dfrac{y^3}{4\left(x+2\right)}+\dfrac{y\left(x+2\right)}{16}\ge\dfrac{y^2}{4}\)
\(\Rightarrow Q+\dfrac{2xy+2x+2y}{16}\ge\dfrac{x^2+y^2}{4}\ge\dfrac{\left(x+y\right)^2}{8}\)
\(\Rightarrow Q\ge\dfrac{\left(x+y\right)^2-\left(x+y\right)}{8}-\dfrac{1}{2}=\dfrac{\left(x+y-4\right)^2+7\left(x+y\right)-16}{8}-\dfrac{1}{2}\)
\(\Rightarrow Q\ge\dfrac{7\left(x+y\right)-16}{8}-\dfrac{1}{2}\ge\dfrac{14\sqrt{xy}-16}{8}-\dfrac{1}{2}=1\)
\(Q_{min}=1\) khi \(x=y=2\)
Áp dụng bất đẳng thức Bunhia dạng phân thức cho 3 số ta có:
\(\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}=\dfrac{2}{2}=1\)
Dấu "=" xảy ra \(\Leftrightarrow\begin{matrix}\dfrac{x}{y+z}=\dfrac{y}{z+x}=\dfrac{z}{x+y}\\x,y,z>0;x+y+z=2\end{matrix}\)
\(\Leftrightarrow x=y=z=\dfrac{2}{3}\)
Áp dụng BĐT Svac-xơ cho 3 số dương có :
\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2.\left(x+y+z\right)}=1\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{2}{3}\)
Vậy Min biểu thức cho là 1 khi \(x=y=z=\dfrac{2}{3}\)
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 \(A_{\min}=22\Leftrightarrow x=y=\frac{1}{2}\)