Let x, y, Z ∈ R +. Prove that (1 + x2) (1 + Y2) (1 + Z2) ≥ 8xyz

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• 1. If ∑ is the outer side of the sphere x ^ 2 + y ^ 2 + Z ^ 2 = 4, then for the surface integral of coordinates ∫ ∫ x ^ 2dxdy, the answer to this problem is 4 π,
• 2. Given: X / 2 = Y / 3 = Z / 4, find (x2 + y2-z2-2xy) / (x2-y2 + z2-2xz) divided by (x2-y2-z2 + 2yz) / (x2 + y2-z2 + 2XY)
• 3. Given x = 1 / 2A + 1, y = 1 / 2A + A, z = A-3, find x2 + 2XY + y2-2xz + z2-2yz The problem is wrong, given x = 1 / 2A + 1, y = 1 / 2A + 2, z = A-3, find the value of x2 + 2XY + y2-2xz + z2-2yz, the answer is 36
• 4. The integral region is ball: x2 + Y2 + Z2
• 5. Who can help to find the triple integral of (XYZ). The region is the first trigram bounded by the sphere x2 + Y2 + Z2 = 1 and the coordinate axis. (where 2 is the square.) I use spherical coordinates to solve the result is 1 / 96, and the answer is 1 / 48
• 6. Find the cylindrical equation parallel to the X axis through the curve 2x ^ 2 + y ^ 2 + Z ^ 2 = 16, x ^ 2 + Z ^ 2-y ^ 2 = 0, I know how to eliminate X and get the equation
• 7. Through the intersection line of two surfaces x ^ 2 + y ^ 2 + 4Z = 1 and x ^ 2 = y ^ 2 + Z ^ 2, the generatrix is parallel to the z-axis,
• 8. Solving cylindrical equation with known bus guide line~ The direction of the generatrix (2,1, - 1) of a cylinder is y ^ 2-4x = 0 and z = 0, and the cylinder equation is written I know the result,
• 9. Let the Quasilinear equation be x + Y-Z = 0, X-Y + Z = 0, and the generatrix be parallel to the straight line x = y = Z
• 10. Find the intersection of three planes x + 3Y + Z = 1, 2x-y-z = 0, - x + 2Y + 2Z = 0
• 11. We know that x, y and Z are all positive numbers, and prove that 33 (LX + 1y + 1z) ≤ 1x2 + 1Y2 + 1z2
• 12. ∫∫ x ^ 2dydz + y ^ 2dzdx + Z ^ 2dxdy ∑ is the lower side of the finite part of the paraboloid z = x ^ 2 + y ^ 2 cut by the plane z = 1
• 13. Calculate the surface integral ∫ (Z ^ 2 + x) dydz zdxdy, where the integral surface is Z = 1 / 2 (x ^ 2 + y ^ 2) between z = 0 and z = 2 Why is the Gauss theorem positive for a closed surface? (the normal vector of a plane is downward, and the angle between it and the z-axis is obtuse. It should be the lower side. It should be negative by using the Gauss theorem,
• 14. Calculate the surface fraction I = ∫∫Σxydydz + 2sinxdxdxdy, where Σ is the lower side of the rotating paraboloid z = x & # 178; + Y & # 178; (0 ≤ Z ≤ 1) Ask for advice
• 15. Surface integral ∫ (2x + Z) dydz + zdxdy integral region: z = x ^ 2 + y ^ 2 (0
• 16. The integral ∫ (x + y) dydz + (y + Z) dzdx + (Z + x) DXDY is calculated by Gauss formula, Where ∑ is the outer part of cylinder x2 + y2 = A2 between 0 ≤ Z ≤ 1
• 17. Find ∫∫∫ sinzdv, where Ω is surrounded by a cone z = root (x ^ 2 + y ^ 2) and a plane y = π
• 18. ∫∫ xdydz + ydzdx + (Z ^ 2-2z) DXDY, where ∑ is the outside of the cone z = root x ^ 2 + y ^ 2 cut by the plane z = 0 and z = 1,
• 19. Calculate I = ∫ (x + | y |) ds, where ∑ is the surface | x | + | y | + | Z | = 1
• 20. ∫ (x ^ 2 + y ^ 2 + Z ^ 2) ds, the integral surface is a closed surface enclosed by x ^ 2 + y ^ 2 + Z ^ 2 = a ^ 2 (x ≥ 0, y ≥ 0) and plane x = 0, y = 0