\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}=\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\dfrac{1}{2xy}\)

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 

\(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}=\dfrac{3\left(2x+y\right)}{16}+\dfrac{3}{2x+y}+\dfrac{5}{16}\left(2x+y\right)\ge2\sqrt{\dfrac{3}{16}.3}+\dfrac{5}{16}.2\sqrt{2xy}=\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\).

Đẳng thức xảy ra khi x = 1; y = 2.

\(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)

\(M=\dfrac{3\left(2x+y\right)}{16}+\dfrac{3}{2x+y}+\dfrac{5\left(2x+y\right)}{16}\ge2\sqrt{\dfrac{9\left(2x+y\right)}{16\left(2x+y\right)}}+\dfrac{5}{16}.2\sqrt{2xy}=\dfrac{11}{4}\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;2\right)\)

Ta có: \(M=\dfrac{2x+y}{xy}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)

\(=\left(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\right)+\dfrac{5}{8}\dfrac{2x+y}{2}\)

Có: \(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\ge2\sqrt{\dfrac{3}{8}\dfrac{2x+y}{2}\dfrac{3}{2x+y}}=\dfrac{3}{2}\)

Dấu '=' xảy ra <=> \(\dfrac{3}{8}\dfrac{2x+y}{2}=\dfrac{3}{2x+y}\)

Có: \(\dfrac{5}{8}\dfrac{2x+y}{2}\ge\dfrac{5}{8}\sqrt{2xy}=\dfrac{5}{4}\)

Dấu '=' xảy ra <=> 2x=y và xy=2

\(\Rightarrow M\ge\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\)

Dấu '=' xảy ra <=> x=1, y=2

Vậy GTNN của M là 11/4 <=> x=1;y=2

Ta có:

\(M=\dfrac{2x+y}{xx}+\dfrac{3}{2x+y}=\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\)

\(=\left(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\right)+\dfrac{5}{8}\dfrac{2x+y}{2}\)

Có: \(\dfrac{3}{8}\dfrac{2x+y}{2}+\dfrac{3}{2x+y}\ge2\sqrt{\dfrac{3}{8}\dfrac{2x+y}{2}\dfrac{3}{2x+y}}=\dfrac{3}{2}\)

Dấu '=' xảy ra \(\Leftrightarrow\dfrac{3}{8}\dfrac{2x+y}{2}=\dfrac{3}{2x+y}\)

Có: \(\dfrac{5}{8}\dfrac{2x+y}{2}\ge\dfrac{5}{8}\sqrt{2xy}=\dfrac{5}{4}\)

Dấu '=' xảy ra \(\Leftrightarrow2x=y,xy=2\)

\(\Rightarrow M\ge\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{11}{4}\)

Dấu '=' xảy ra \(\Leftrightarrow x=1,y=2\)

Vậy GTNN của M là \(\dfrac{11}{4}\Leftrightarrow x=1,y=2\)

Lời giải:

Áp dụng BĐT Bunhiacopxky:

$(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy})(x^2+y^2+2xy)\geq (1+1+2)^2=16$

$\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}\geq \frac{16}{(x+y)^2}=16$

Áp dụng BĐT AM-GM:

$xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}$

$\Rightarrow \frac{2}{xy}\geq 8$

Cộng 2 BĐT trên lại:

$P\geq 16+8=24$

Vậy $P_{\min}=24$ khi $x=y=\frac{1}{2}$

Lời giải:

Áp dụng BĐT Bunhiacopxky:

$(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy})(x^2+y^2+2xy)\geq (1+1+2)^2=16$

$\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}\geq \frac{16}{(x+y)^2}=16$

Áp dụng BĐT AM-GM:

$xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}$

$\Rightarrow \frac{2}{xy}\geq 8$

Cộng 2 BĐT trên lại:

$P\geq 16+8=24$

Vậy $P_{\min}=24$ khi $x=y=\frac{1}{2}$

*cách này đơn giản hơn

Vì x,y>0. theo AM-GM:

\(\dfrac{1}{x^2}\)+\(\dfrac{1}{y^2}\) ≥\(\dfrac{2}{xy}\) => P≥\(\dfrac{6}{xy}\)

ta có: \(x^2\)+\(y^2\)≥ 2xy <=> (x+y)\(^2\)≥4xy <=> xy≤\(\dfrac{\left(x+y\right)^2}{4}\)=\(\dfrac{1}{4}\)

<=> \(\dfrac{6}{xy}\)≥\(\)24 hay P≥24

dấu = xảy ra khi: x=y=\(\dfrac{1}{2}\)

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 \(x=y=z=\dfrac{2}{3}\)

\(xy\ge2\left(y-1\right)\ge0\Rightarrow x\ge\dfrac{2\left(y-1\right)}{y}\ge0\)

\(\Rightarrow M\ge\dfrac{\dfrac{4\left(y-1\right)^2}{y^2}+4}{y^2+1}=4.\dfrac{\left(y-1\right)^2+y^2}{y^2\left(y^2+1\right)}\)

\(\dfrac{M}{4}\ge\dfrac{2y^2-2y+1}{y^4+y^2}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{\left(2-y\right)\left(y^3+2y^2-3y+2\right)}{4\left(y^4+y^2\right)}+\dfrac{1}{4}\ge\dfrac{1}{4}\)

\(\Rightarrow M\ge1\)

Dấu "=" xảy ra khi \(y=2;x=1\)

Ta có: \(Q=\dfrac{2}{x^2+y^2}+\dfrac{3}{xy}=\dfrac{2}{x^2+y^2}+\dfrac{6}{2xy}=\dfrac{2}{x^2+y^2}+\dfrac{2}{2xy}+\dfrac{4}{2xy}\)

Áp dụng BĐT phụ: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\Rightarrow2\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)\ge2\left(\dfrac{4}{x^2+2xy+y^2}\right)=2\left[\dfrac{4}{\left(x+y\right)^2}\right]=2.\dfrac{4}{4}=2\)

Dấu "=" xảy ra khi x=y=1

Áp dụng BĐT phụ: \(ab\le\dfrac{\left(a+b\right)^2}{4}\)

\(\Rightarrow xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{2^2}{4}=1\)

Dấu"=" xảy ra khi x=y=1

\(\Rightarrow2xy\le2.1=2\)

\(\Rightarrow\dfrac{4}{2xy}\ge\dfrac{4}{2}=2\)

\(\Rightarrow Q=\dfrac{2}{x^2+y^2}+\dfrac{2}{2xy}+\dfrac{4}{2xy}=\dfrac{2}{x^2+y^2}+\dfrac{3}{xy}\ge2+2=4\)

Dấu"=" xảy ra khi x=y=1

Em kiểm tra đề là \(\dfrac{y^2}{4}\) hay \(\dfrac{y^4}{4}\)

Nếu đề đúng là \(\dfrac{y^4}{4}\) thì có thể coi như là không giải được

\(2x^2+\dfrac{1}{x^2}+\dfrac{y^2}{4}=4\Leftrightarrow\left(x^2+\dfrac{1}{x^2}-2\right)+\left(x^2-xy+\dfrac{y^2}{4}\right)+xy=2\)

\(\Leftrightarrow2=\left(x-\dfrac{1}{x}\right)^2+\left(x-\dfrac{y}{2}\right)^2+xy\ge xy\)

\(\Rightarrow P_{max}=2023\) khi \(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\x-\dfrac{y}{2}=0\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(-1;-2\right);\left(1;2\right)\)

\(2x^2+\dfrac{1}{x^2}+\dfrac{y^2}{4}=4\Leftrightarrow\left(x^2+\dfrac{1}{x^2}-2\right)+\left(x^2+xy+\dfrac{y^2}{4}\right)-xy=2\)

\(\Rightarrow2=\left(x-\dfrac{1}{x}\right)^2+\left(x+\dfrac{y}{2}\right)^2-xy\ge-xy\)

\(\Rightarrow xy\ge-2\Rightarrow P\ge2019\)

\(P_{min}=2019\) khi \(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\x+\dfrac{y}{2}=0\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(-1;2\right);\left(1;-2\right)\)