# 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<= 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.

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

### 1 Answer

- Nino_opsLv 61 decade agoFavorite Answer
Im not sure abt what r u asking in the first Q

Why its given on a b anc and asking for x y and z ? ?

are ther more details like a+b-c > 0 ? ?

or a b and c are sides of a triangel ? ?

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

Use the AG GM inequality on

1). a^2 and b^4 [ a^2 ≥ 0 , b^4 ≥ 0]

(a^2 + b^4)/2 ≥ sqrt(a^2 b^4)

(a^2 + b^4)/2 ≥ ab^2

2). b^4 and c^6 [ b^4 ≥ 0 , c^6 ≥ 0]

(b^4 + c^6)/2 ≥ sqrt(b^4 c^6)

(b^4 + c^6)/2 ≥ b^2 c^3

3). c^6 and d^8 [ c^6 0 , d^8 ≥ 0]

(c^6 + d^8)/2 ≥ sqrt(c^6 d^8)

(c^6 + d^8(/2 ≥ c^3 d^4

4). d^8 and a^2

(a^2 + d^8)/2 ≥ sqrt(a^2 d^78)

(a^2 + d^8)/2 ≥ ad^4

Now add 1 2 3 4 together

(a^2 + b^4 + b^4 + c^6 + c^6 + d^8 + d^8 + a^2)/2 ≥ ab^2 + b^2 c^3 + c^3 d^4 + ad^4

(a^2 + b^4 + c^6 + d^8) ≥ ab^2 + b^2 c^3 + c^3 d^4 + ad^4

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

Let x y and z be any real positive non zero numbers.

F(x , y) = x^3 + y^3 – x^2y – y^2x

F(x , y) = x^2 (x – y) - y^2(x – y)

F(x , y) = (x – y)(x^3 – y^3)

F(x , y) = (x – y)(x – y)(x^2 + y^2 + xy)

F(x , y) = (x – y)^2 (x^2 + y^2 + xy)

(x^2 + y^2 + xy) is always positive

(x – y)^2 is always positive but it is 0 when x = y

Therefore;

F(x , y) ≥ 0

(x – y)^2 (x^2 + y^2 + xy) ≥ 0

x^3 + y^3 – x^2y – y^2x ≥ 0

Since we prooved this for any x y z (x,y,x positive non-zero numbers)

we can use it on a b and c

a^3 + b^3 - a^2b - b^2a ≥ 0

a^3 + c^3 - a^2c- c^2a ≥ 0

c^3 + b^3 - c^2b - b^2c ≥ 0

adding them together;

2 (a^3 + b^3 + c^3) - a^2b - b^2a - a^2c- c^2a - c^2b - b^2c ≥ 0

2(a^3 + b^3 + c^3) ≥ a^2b +b^2a + a^2c + c^2a + c^2b + b^2c