Do \(x-y=\dfrac{x+y}{\sqrt{xy}}>0\Rightarrow x>y\)

Khi đó:

\(\sqrt{xy}\left(x-y\right)=x+y\Rightarrow xy\left(x-y\right)^2=\left(x+y\right)^2\)

\(\Rightarrow xy\left[\left(x+y\right)^2-4xy\right]=\left(x+y\right)^2\)

\(\Rightarrow\left(xy-1\right)\left(x+y\right)^2=4x^2y^2\)

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

Do vế trái dương nên vế phải dương \(\Rightarrow xy-1>0\)

\(\Rightarrow\left(x+y\right)^2=\dfrac{4x^2y^2-4+4}{xy-1}=4xy+4+\dfrac{4}{xy-1}=4\left(xy-1\right)+\dfrac{4}{xy-1}+8\)

\(\ge2\sqrt{4\left(xy-1\right).\dfrac{4}{xy-1}}+8=16\)

\(\Rightarrow x+y\ge4\)

\(P_{min}=4\) khi \(\left(x;y\right)=\left(2+\sqrt{2};2-\sqrt{2}\right)\)

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 \(x=y=\dfrac{1}{2}\)

Áp dụng BĐT Cauchy cho cặp số dương \(\dfrac{1}{\left(z+x\right)};\dfrac{1}{\left(z+y\right)}\)

\(\dfrac{1}{\left(z+x\right)}+\dfrac{1}{\left(z+y\right)}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\)

\(\Rightarrow\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\left(1\right)\)

Tương tự ta được

\(\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}\le\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}\left(2\right)\)

\(\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\left(3\right)\)

\(\left(1\right)+\left(2\right)+\left(3\right)\) ta được :

\(P=\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}+\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}+\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\)

\(\Rightarrow P\le2\left(x+y+z\right)=2.3=6\)

\(\Rightarrow GTLN\left(P\right)=6\left(tạix=y=z=1\right)\)

CHÚC BẠN HỌC TỐT

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

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

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

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

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

\(\left|xy\right|+\left|yz\right|+\left|zx\right|\)