Prove inequality (U.S. Math Olympiad, 1997)?

Question

Prove that for all positive real numbers a,b,c it holds true that

1/(a³ + b³ + abc) + 1/(b³ + c³ + abc)

+ 1/(c³ + a³ + abc) ≤ 1/(abc).

Answer 1

Factorize a^3+b^3:

a^3+b^3 = (a^2-ab+b^2)(a+b) =

(a^2-2ab+b^2 + ab) (a+b) =

((a-b)^2 + ab) (a+b) >= ab (a+b).

Then a^3+b^3+ abc  >= ab (a+b+c).

Similarly,

a^3+c^3+ abc  >= ac (a+b+c),

b^3+c^3+ abc  >= bc (a+b+c).

Hence the left side is less than or equal to

(1/(a+b+c)) (1/(ab) + 1/(ac) + 1/(bc)) = 1/(abc).

Equality is possible when a=b=c.

Answer 2

By the AM-GM inequality a³+b³+abc > 3(a³b³abc)^(1/3) = (a^4b^4c)^(1/3) = 3ab(abc)^(1/3)

Similarly:

a³+c³+abc > 3ac(abc)^(1/3) and b³+c³+abc > 3bc(abc)^(1/3)

So 1/(a³+b³+abc) <  1/3ab(abc)^(1/3) = c/3(abc)^(4/3)

And 1/(a³+c³+abc) < 1/3ac(abc)^(1/3) = b/3(abc)^(4/3)

And 1/(b³+c³+abc) < 1/3bc(abc)^(1/3) = a/3(abc)^(4/3)

So

1/(a³+b³+abc) + 1/(a³+c³+abc) + 1/(b³+c³+abc)

< c/3(abc)^(4/3) + b/3(abc)^(4/3) + a/3(abc)^(4/3)

= (a+b+c)/3(abc)^(4/3)

Now I thought I could use the AM-GM inequality here but it goes in the wrong direction. The AM-GM inequality (a+b+c)/3 > (abc)^(1/3)

Maybe this will give someone some insight. I'm going back and starting again.

Ok here goes second attempt.

1/(a³+b³+abc) = 1/(3a³b³) * 3/(1/a³+1/b³+c/(a²b²))

By the AM-HM inequality:

3/(1/a³+1/b³+a²b²/c) < (a³+b³+a²b²/c)/3

So

1/(a³+b³+abc)

< 1/(3a³b³) * (a³+b³+a²b²/c)/3

= (a³+b³+a²b²/c)/(9a³b³)

= (a³c³+b³c³+a²b²c²)/(9a³b³c³)

Similarly

1/(a³+c³+abc) < (a³b³+b³c³+a²b²c²)/(9a³b³c³)

1/(a³+b³+abc) < (a³c³+a³c³+a²b²c²)/(9a³b³c³)

So

1/(a³+b³+abc) + 1/(a³+c³+abc) + 1/(a³+b³+abc)

< (a³c³+b³c³+a²b²c²)/(9a³b³c³) + (a³b³+b³c³+a²b²c²)/(9a³b³c³) + (a³c³+a³c³+a²b²c²)/(9a³b³c³)

= (2a³b³+2a³c³+2b³c³+3a²b²c²)/(9a³b³c³)

= 2/(9a³) + 2/(9b³) + 2/(9c³) + 1/(3abc)

Which again is still too big.

All the variations of AM-GM-HM inequality that I have used end up as either:

LHS < (a+b+c)/3(abc)^(4/3)

or

LHS < 2/(9a³) + 2/(9b³) + 2/(9c³) + 1/(3abc)