We have: \(\left|7x-5y\right|\ge0\); \(\left|2z-3x\right|\ge0\) and \(\left|xy+yz+zx-2000\right|\ge0\)

=> \(P\ge0+0+0\ge0\)

\(P=0\) only when \(7x-5y=2z-3x=xy+yz+zx-2000=0\)

=> \(7x=5y;2z=3x;xy+yz+zx=2000\)

=> \(\dfrac{x}{5}=\dfrac{y}{7}\); \(\dfrac{z}{3}=\dfrac{x}{2}\) and \(xy+yz+zx=2000\)

=> \(\dfrac{x}{10}=\dfrac{y}{14}=\dfrac{z}{15}\) and \(xy+yz+zx=2000\)

Put \(\dfrac{x}{10}=\dfrac{y}{14}=\dfrac{z}{15}=k\)

We have: \(k^2=\dfrac{x}{10}\cdot\dfrac{y}{14}=\dfrac{y}{14}\cdot\dfrac{z}{15}=\dfrac{z}{15}\cdot\dfrac{x}{10}=\dfrac{xy+yz+zx}{10\cdot14+14\cdot15+15\cdot10}=\dfrac{2000}{140+210+150}\)

\(=\dfrac{2000}{500}=4\)

=> \(k=\pm2\)

=> \(\left(x;y;z\right)=\left(20;28;30\right)\) and  \(\left(-20;-28;-30\right)\)

So \(minP=0\) only when (x;y;z) = (20;28;30) and (-20;-28;-30)

We have: |7x−5y|≥0; |2z−3x|≥0 and |xy+yz+zx−2000|≥0

=> P≥0+0+0≥0

P=0

 only when 7x−5y=2z−3x=xy+yz+zx−2000=0

=> 7x=5y;2z=3x;xy+yz+zx=2000

=> x5=y7

; z3=x2 and xy+yz+zx=2000

=> x10=y14=z15

 and xy+yz+zx=2000

Put x10=y14=z15=k

We have: k2=x10⋅y14=y14⋅z15=z15⋅x10=xy+yz+zx10⋅14+14⋅15+15⋅10=2000140+210+150

=2000500=4

=> k=±2

=> (x;y;z)=(20;28;30)

 and  (−20;−28;−30)

So minP=0

 only when (x;y;z) = (20;28;30) and (-20;-28;-30)