a: f(0)=-4,5

f(-1)=4,5

Làm như vậy hả

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\(x^3+y^3+3xy\le1\Leftrightarrow\left(x+y\right)^3-1-3xy\left(x+y\right)+3xy\le0\)

\(\Leftrightarrow\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\le0\)

\(\Leftrightarrow\left(x+y-1\right)\left(x^2+y^2-xy+x+y+1\right)\le0\)

Do \(x^2+y^2-xy+x+y+1=\left(x-\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+x+y+1>0\)

\(\Rightarrow x+y-1\le0\Rightarrow x+y\le1\)

\(\Rightarrow P=\left(x+\dfrac{1}{4x}\right)+\left(y+\dfrac{1}{4y}\right)+\dfrac{3}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

\(\Rightarrow P\ge2\sqrt{\dfrac{x}{4x}}+2\sqrt{\dfrac{y}{4y}}+\dfrac{3}{4}.\dfrac{4}{x+y}\ge2+\dfrac{3}{4}.\dfrac{4}{1}=5\)

\(P_{min}=5\) khi \(x=y=\dfrac{1}{2}\)

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

\(=\frac{1}{x^2+y^2}+\frac{1}{2xy}+4xy+\frac{3}{2xy}\)

\(\ge\frac{4}{x^2+y^2+2xy}+4xy+\frac{1}{4xy}+\frac{5}{4xy}\)

\(\ge\frac{4}{x^2+y^2+2xy}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{5}{4xy}\)

Ta có BĐT phụ: \(\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\)(đúng )

Dấu "=" xảy ra <=> x=y

\(\Rightarrow P\ge\frac{4}{\left(x+y\right)^2}+2+\frac{5}{\left(x+y\right)^2}\)

\(\ge\frac{4}{1}+2+\frac{5}{1}=11\)

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

Vậy Min P =11 \(\Leftrightarrow x=y=\frac{1}{2}\)

\(\dfrac{1}{x^2+2}-\dfrac{1}{xy+2}+\dfrac{1}{y^2+2}-\dfrac{1}{xy+2}=0\)

\(\Leftrightarrow\dfrac{xy-x^2}{\left(x^2+2\right)\left(xy+2\right)}+\dfrac{xy-y^2}{\left(y^2+2\right)\left(xy+2\right)}=0\)

\(\Leftrightarrow\dfrac{x-y}{xy+2}\left(\dfrac{y}{y^2+2}-\dfrac{x}{x^2+2}\right)=0\)

\(\Leftrightarrow\left(\dfrac{x-y}{xy+2}\right)\left(\dfrac{x^2y+2y-xy^2-2x}{\left(x^2+2\right)\left(y^2+2\right)}\right)=0\)

\(\Leftrightarrow\dfrac{\left(x-y\right)^2\left(xy-2\right)}{\left(xy+2\right)\left(x^2+2\right)\left(y^2+2\right)}=0\)

\(\Leftrightarrow xy=2\) (do x;y phân biệt)

\(\Rightarrow P=\dfrac{2}{xy+2}+\dfrac{2}{xy+2}=\dfrac{4}{xy+2}=\dfrac{4}{2+2}=1\)