\(P=\frac{2x^2+y^2-2xy}{xy}=\frac{2x}{y}+\frac{y}{x}-2=\frac{7x}{4y}+\left(\frac{x}{4y}+\frac{y}{x}-2\right)\)

Áp dụng BĐT Cô - Si cho các số dương :
\(\frac{x}{4y}+\frac{y}{x}\ge2\sqrt{\frac{x}{4y}.\frac{y}{x}}=1\)

\(\frac{7x}{4y}\ge\frac{7.2y}{4y}=\frac{7}{2}\) do \(x\ge2y\)

Do đó : \(P\ge\frac{7}{2}+1-2=\frac{5}{2}\)

Vậy \(P_{min}=\frac{5}{2}\) khi x\(=2y\)

Chúc bạn học tốt !!!

Từ điều kiện bài toán ta có

\(\hept{\begin{cases}\frac{x}{y}\ge1\\x-y\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{x}{y}\ge1\\x^2-2xy+y^2\ge0\end{cases}}\)

Thế vào ta được

\(P=\frac{2x^2+y^2-2xy}{xy}\ge\frac{x^2}{xy}=\frac{x}{y}\ge1\)

Dấu = xảy ra khi x = y

Ta có:

\(A=\frac{2x^2+y^2-2xy}{xy}=\frac{\left(x^2-4xy+4y^2\right)+x^2+2xy-3y^2}{xy}=\frac{\left(x-2y\right)^2+x^2+2xy-3y^2}{xy}\)

\(=\frac{\left(x-2y\right)^2}{xy}+\frac{x}{y}+2+\frac{-3y}{x}\ge0+2+2+\frac{-3}{2}=\frac{5}{2}\)

Vậy minA = \(\frac{5}{2}\)khi x = 2y.

Bài 1:
\(P=\frac{2x^2+y^2-2xy}{xy}=\frac{2x}{y}+\frac{y}{x}-2=\frac{7x}{4y}+(\frac{x}{4y}+\frac{y}{x})-2\)

Áp dụng BĐT Cô-si cho các số dương:

\(\frac{x}{4y}+\frac{y}{x}\geq 2\sqrt{\frac{x}{4y}.\frac{y}{x}}=1\)

\(\frac{7x}{4y}\geq \frac{7.2y}{4y}=\frac{7}{2}\) do $x\geq 2y$

Do đó: \(P\geq \frac{7}{2}+1-2=\frac{5}{2}\)

Vậy $P_{\min}=\frac{5}{2}$ khi $x=2y$

Bài 2:
\(P=\frac{x^2+y^2}{x^2y^2}+\frac{x^2y^2}{x^2+y^2}=\frac{x^2+y^2}{\frac{1}{4}}+\frac{1}{4(x^2+y^2)}=4(x^2+y^2)+\frac{1}{4(x^2+y^2)}\)

Áp dụng BĐT Cô-si :

\(\frac{x^2+y^2}{4}+\frac{1}{4(x^2+y^2)}\geq 2\sqrt{\frac{x^2+y^2}{4}.\frac{1}{4(x^2+y^2)}}=\frac{1}{2}(1)\)

\(x^2+y^2\geq 2\sqrt{x^2y^2}=2|xy|=2.\frac{1}{2}=1\)

\(\Rightarrow \frac{15(x^2+y^2)}{4}\geq \frac{15}{4}(2)\)

Lấy \((1)+(2)\Rightarrow P\geq \frac{15}{4}+\frac{1}{2}=\frac{17}{4}\)

Vậy \(P_{\min}=\frac{17}{4}\Leftrightarrow x=y=\frac{1}{\sqrt{2}}\)

Bài 3:

Có: \(a^2+\frac{18}{\sqrt{a}}=\frac{7}{16}a^2+\frac{9}{16}a^2+\frac{18}{\sqrt{a}}\)

Áp dụng BĐT Cô-si:

\(\frac{9}{16}a^2+\frac{18}{\sqrt{a}}\geq 2\sqrt{\frac{9}{16}a^2.\frac{18}{\sqrt{a}}}=\frac{9}{2}\sqrt{2a\sqrt{a}}\geq \frac{9}{2}\sqrt{2.4\sqrt{4}}=18(1)\) do $a\geq 4$

\(\frac{7}{16}a^2\geq \frac{7}{16}.4^2=7(2)\) do $a\geq 4$

Lấy \((1)+(2)\Rightarrow a^2+\frac{18}{\sqrt{a}}\geq 7+18=25\) (đpcm)

Dấu "=" xảy ra khi $a=4$

Các bất đẳng thức đúng : \(ab\le\frac{\left(a+b\right)^2}{4};\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

Áp dụng ta được :

\(A=\frac{1}{x^2+y^2}+\frac{2}{xy}\)

\(=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{3}{2xy}\)

Ta có :

\(\frac{1}{x^2+y^2}+\frac{1}{2xy}\ge\frac{4}{x^2+y^2+2xy}=\frac{4}{\left(x+y\right)^2}\ge4\)

\(\frac{3}{2xy}\ge\frac{3}{2.\frac{\left(x+y\right)^2}{4}}=\frac{3}{2.\frac{1}{4}}=6\)

\(\Rightarrow A=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{3}{2xy}\ge4+6=10\)

Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)

Vậy \(A_{min}=10\) tại \(x=y=\frac{1}{2}\)

thangwd hdashdfjdfishjdf

Áp dụng nè : \(\frac{2}{x^2+y^2}+\frac{2}{2xy}\ge\frac{8}{\left(x+y\right)^2}\ge\frac{1}{2}\)

khó was

BẠn cố gắng áp dụng chọn điểm rơi và bđt nè :\(\frac{2}{x^2+y^2} +\frac{2}{2xy}\ge\frac{8}{\left(x+y\right)^2}\)

Nếu ko lm đc tiwps vui lòng cmt

\(y\ge xy+1\ge2\sqrt{xy}\Rightarrow\sqrt{\dfrac{y}{x}}\ge2\Rightarrow\dfrac{y}{x}\ge4\)

\(Q=\dfrac{1-\dfrac{2y}{x}+2\left(\dfrac{y}{x}\right)^2}{\dfrac{y}{x}+\left(\dfrac{y}{x}\right)^2}\)

Đặt \(\dfrac{y}{x}=a\ge4\)

\(Q=\dfrac{2a^2-2a+1}{a^2+a}=\dfrac{2a^2-2a+1}{a^2+a}-\dfrac{5}{4}+\dfrac{5}{4}=\dfrac{\left(a-4\right)\left(3a-1\right)}{4\left(a^2+1\right)}+\dfrac{5}{4}\ge\dfrac{5}{4}\)

\(Q_{min}=\dfrac{5}{4}\) khi \(a=4\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le\dfrac{1}{4}\)

Đặt \(\dfrac{x}{y}=a\Rightarrow0< a\le\dfrac{1}{4}\)

\(P=\dfrac{\left(\dfrac{x}{y}\right)^2-\dfrac{2x}{y}+2}{\dfrac{x}{y}+1}=\dfrac{a^2-2a+2}{a+1}=\dfrac{4a^2-8a+8}{4\left(a+1\right)}=\dfrac{4a^2-13a+3+5\left(a+1\right)}{4\left(a+1\right)}\)

\(P=\dfrac{5}{4}+\dfrac{\left(1-4a\right)\left(3-a\right)}{4\left(a+1\right)}\ge\dfrac{5}{4}\)

Dấu "=" xảy ra khi \(a=\dfrac{1}{4}\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)