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

\(\Rightarrow\sqrt{x}+\sqrt{y}+2\ge4\)

\(\Rightarrow2\le\sqrt{x}+\sqrt{y}\le\sqrt{2\left(x+y\right)}\Rightarrow x+y\ge2\)

\(\Rightarrow P\ge\dfrac{\left(x+y\right)^2}{x+y}=x+y\ge2\)

Dấu "=" xảy ra khi \(x=y=1\)

Trước hết áp dụng BĐT: \(ab\le\dfrac{1}{4}\left(a+b\right)^2\)

Ta có: \(\left(\sqrt{x}+1\right)\left(\sqrt{y}+1\right)\le\dfrac{1}{4}\left(\sqrt{x}+1+\sqrt{y}+1\right)^2\)

Mà \(\left(\sqrt{x}+1\right)\left(\sqrt{y}+1\right)\ge4\Rightarrow\dfrac{1}{4}\left(\sqrt{x}+\sqrt{y}+2\right)^2\ge4\)

\(\Rightarrow\left(\sqrt{x}+\sqrt{y}+2\right)^2\ge4^2\)

\(\Rightarrow\sqrt{x}+\sqrt{y}+2\ge4\)

\(\Rightarrow\sqrt{x}+\sqrt{y}\ge2\)

Lại áp dụng tiếp: \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\Rightarrow a+b\le\sqrt{2\left(a^2+b^2\right)}\)

Ta được: \(\sqrt{x}+\sqrt{y}\le\sqrt{2\left(x+y\right)}\)

\(\Rightarrow\sqrt{2\left(x+y\right)}\ge\sqrt{x}+\sqrt{y}\ge2\)

Bình phương lên: \(2\left(x+y\right)\ge4\Rightarrow x+y\ge2\)

Phần cuối chắc là hoàn toàn cơ bản rồi

Biến đổi từ giả thiết

\(x^3+y^3+6xy\le8\)

\(\Leftrightarrow...\Leftrightarrow\left(x+y-2\right)\left(x^2-xy+y^2+2x+2y+4\right)\le0\)

\(\Leftrightarrow x+y-2\le0\)

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

\(\Leftrightarrow x+y\le2\)

Và áp dụng các bđt \(\frac{1}{2ab}\ge\frac{2}{\left(a+b\right)^2}\)

                                 \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\left(a;b>0\right)\)

Khi đó \(P=\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\left(\frac{1}{ab}+ab\right)+\frac{3}{2ab}\)

               \(\ge\frac{4}{a^2+b^2+2ab}+2+\frac{6}{\left(a+b\right)^2}\)

                 \(=\frac{4}{\left(a+b\right)^2}+2+\frac{6}{\left(a+b\right)^2}\ge\frac{9}{2}\)

Dấu "=" <=> a= b = 1

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 \(x=y=\dfrac{1}{2}\)