# More Inequality Proofs...?

Thanks for help with previous problems, now I'm having some problems with these harder questions. Note, >= means bigger than or equal to, and <= means smaller than or equal to.
1) Suppose a^3 + b^3 + c^3 >= 3abc, x + y + z >= 3[cubedrt(xyz)], (1+x)(1+y)(1+z) = 8
Therefore prove xyz<=...
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Thanks for help with previous problems, now I'm having some problems with these harder questions. Note, >= means bigger than or equal to, and <= means smaller than or equal to.

1) Suppose a^3 + b^3 + c^3 >= 3abc, x + y + z >= 3[cubedrt(xyz)], (1+x)(1+y)(1+z) = 8

Therefore prove xyz<= 1

2) Prove that a^2 + b^4 + c^6 + d^8 >= ab^2 + b^2c^3 + c^3d^4 + d^4a for all real numbers.

Included hint is: think of ab^2 as sqrt(a^2b^4) and use the arithmetic/geometric mean theorem.

3) Prove that 2(a^3 + b^3 + c^3) >= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 for positive real numbers.

Included hint is: this is the same as proving 2(a^3 + b^3 + c^3) – (a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2) >= 0. Then break it up. Start by showing that a^3 + b^3 – a^2b – ab^2 >=0, by using and rearranging factors. Then do same for b^3 + c^3, etc.

Thankyou again for all help in advance.

1) Suppose a^3 + b^3 + c^3 >= 3abc, x + y + z >= 3[cubedrt(xyz)], (1+x)(1+y)(1+z) = 8

Therefore prove xyz<= 1

2) Prove that a^2 + b^4 + c^6 + d^8 >= ab^2 + b^2c^3 + c^3d^4 + d^4a for all real numbers.

Included hint is: think of ab^2 as sqrt(a^2b^4) and use the arithmetic/geometric mean theorem.

3) Prove that 2(a^3 + b^3 + c^3) >= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 for positive real numbers.

Included hint is: this is the same as proving 2(a^3 + b^3 + c^3) – (a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2) >= 0. Then break it up. Start by showing that a^3 + b^3 – a^2b – ab^2 >=0, by using and rearranging factors. Then do same for b^3 + c^3, etc.

Thankyou again for all help in advance.

Update:
Umm no more details, but it's ok I solved the first question anyway. Thank you greatly for the solutions to the other 2!

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