Ta có:

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

đặt x+y=a và xy=b

\(\Rightarrow a^2=b\left(a^2-4b\right)\Rightarrow a^2=a^2b-4b^2\Rightarrow4b^2=a^2\left(b-1\right)\Rightarrow\frac{4b^2}{b-1}=a^2\)

Lại có:

\(\frac{b^2}{b-1}=\frac{b^2-1+1}{b-1}=b+1+\frac{1}{b-1}=b-1+\frac{1}{b-1}+2\ge2+2=4\)

\(\Rightarrow\frac{4b^2}{b-1}\ge16\Rightarrow a^2\ge16\Rightarrow a\ge4\Rightarrow x+y\ge4\)

Dấu bằng xảy ra khi \(x=2+\sqrt{2},y=2-\sqrt{2}\)

Có cách khác nè:

P=x4(x−1)3+y4(y−1)3≥2√x4y4(x−1)3(y−1)3x4(x−1)3+y4(y−1)3≥2x4y4(x−1)3(y−1)3

⇒P≥2x2y2√(x−1)3(y−1)3=2.x2x−1.y2y−1.1√(x−1)(y−1)⇒P≥2x2y2(x−1)3(y−1)3=2.x2x−1.y2y−1.1(x−1)(y−1)

Ta dễ dàng chứng minh được a2a−1≥4a2a−1≥4

⇒P≥2.4.4.1√(x−1)(y−1)≥32.1x−1+y−12≥32⇒P≥2.4.4.1(x−1)(y−1)≥32.1x−1+y−12≥32

Dấu "=" khi x=y=2

x4(x−1)3+16(x−1)≥8.x2(x−1)x4(x−1)3+16(x−1)≥8.x2(x−1)

Tương tự và cộng hai BĐT lại : 

p+16(x−1)+16(y−1)≥8.(x2x−1+y2y−1)p+16(x−1)+16(y−1)≥8.(x2x−1+y2y−1)

Ta xét A=x2x−1+y2y−1A=x2x−1+y2y−1

Đặt x - 1 = a và y - 1 = b, ta có A=(a+1)2a+(b+1)2b=a+2+1a+b+2+1b≥(a+b)+4a+b+4≥2√4+4=8⇒A≥8A=(a+1)2a+(b+1)2b=a+2+1a+b+2+1b≥(a+b)+4a+b+4≥24+4=8⇒A≥8

Do đó P≥8A−16(x+y)+32≥8.8−16.4+32=32P≥8A−16(x+y)+32≥8.8−16.4+32=32

Min P = 32 <=> x = y = 2 

Thay \(xy+yz+zx=5\) vào P, ta có:

\(P=\frac{3x+3y+2z}{\sqrt{6\left(x+y\right)\left(x+z\right)}+\sqrt{6\left(y+z\right)\left(y+x\right)}+\sqrt{\left(z+x\right)\left(z+y\right)}}\)

Áp dụng bất đẳng thức Cô-si, ta có:

\(\sqrt{6\left(x+y\right)\left(x+z\right)}\le\frac{3\left(x+y\right)+2\left(x+z\right)}{2}\)

\(\sqrt{6\left(y+z\right)\left(y+x\right)}\le\frac{3\left(y+x\right)+2\left(y+z\right)}{2}\)

\(\sqrt{\left(z+x\right)\left(z+y\right)}\le\frac{\left(z+x\right)+\left(z+y\right)}{2}\)

Cộng vế theo vế các bất đẳng thức cùng chiều, ta đươc:

\(\sqrt{6\left(x+y\right)\left(x+z\right)}+\sqrt{6\left(y+z\right)\left(y+x\right)}+\sqrt{\left(z+x\right)\left(z+y\right)}\le\frac{9}{2}x+\frac{9}{2}y+3z\)

\(\Rightarrow P\ge\frac{3x+3y+2z}{\frac{9}{2}x+\frac{9}{2}y+3z}=\frac{3x+3y+2z}{\frac{3}{2}\left(3x+3y+2z\right)}=\frac{2}{3}\)

Dấu "=" khi \(\hept{\begin{cases}3\left(x+y\right)=2\left(y+z\right)=2\left(z+x\right)\\z+y=z+x\\xy+yz+zx=5\end{cases}\Leftrightarrow\hept{\begin{cases}x=y=1\\z=2\end{cases}}}\)

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 :

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

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

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

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 \(\left(x;y\right)=\left(2+\sqrt{2};2-\sqrt{2}\right)\)