1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:

\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).

Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).

2.

\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)

Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)

\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )

\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)

\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)

Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)

3. Chia 2 vế giả thiết cho \(x^2y^2\)

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

\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)

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

Vì x,y là số thực dương nên theo BĐT Cosi ta có:

\(x+y\ge2\sqrt{xy}\) Dấu "=" xảy ra <=> x=y hay x+x+x²=15 => x=y=3

GT: x+y+xy=15 => xy=15-(x+y)

Do đó: \(P=x^2+y^2=\left(x+y\right)^2-2xy=\left(x+y\right)^2-30+2\left(x+y\right)\ge\left(2\sqrt{xy}\right)^2-30+2\cdot2\sqrt{xy}\)

Dấu "=" xảy ra <=> x=y=3

Vậy \(min_P=4\cdot3^2-30+4\cdot3=18\Leftrightarrow x=y=3\)

ta có x>=2y suy ra x-2y>=0

m=x^2/xy+y^2/xy điều kiện x,y khác 0

M=x/y+y/x

2M=2x/y+2y/x

2M=2.x/y+(-x+2y+x)/x

2m=2.(x-2y)/y+2.2y/x-(x-2y)/x+x/x

2m=2(x-2y)/y-(x-2y)/x+5

vì x-2y>=0=>2(x-2y)/y-(x-2y)/x+5>=5

2M>=5

2M>5/2

vậy M=5/2

chưa chắc đã đúg đôu đúg tk mk nha

Đặt \(\frac{x}{y}=a\)

Vì \(x\ge2y>0\Rightarrow a\ge2\)

Khi đó \(P=\frac{x}{y}+\frac{y}{x}=a+\frac{1}{a}=\left(\frac{1}{a}+\frac{a}{4}\right)+\frac{3a}{4}\ge2\sqrt{\frac{1}{a}.\frac{a}{4}}+\frac{3a}{4}\ge1+\frac{3}{2}=\frac{5}{2}\)

Dấu " \(=\)" xảy ra \(\Leftrightarrow\)\(a=2\Leftrightarrow x=2y>0\)

Từ đề bài \(\Rightarrow4x^2+4y^2+4xy-24x-24y+44=0\)

\(\Leftrightarrow\left(2x+y\right)^2-24x-12y+36+3y^2-12y+12-4=0\)

\(\Leftrightarrow\left(2x+y-6\right)^2+3\left(y-2\right)^2-4=0\)

\(\Leftrightarrow\left(2x+y-6\right)^2=4-3\left(y-2\right)^2\le4\forall x;y\)

\(\Leftrightarrow-2\le2x+y-6\le2\Rightarrow4\le2x+y\le8\)

Do đó \(4\le P\le8\)