let $a,b,c$ are postive numbers, show that
$$(a^3+b^3+c^3)(ab+bc+ac)\ge 6abc(a^2+b^2+c^2-ab-bc-ac)$$
my try: let $$a+b+c=p,ab+bc+ac=q,abc=r$$ and the $$a^3+b^3+c^3=(a+b+c)^3-3(ab+bc+ac)(a+b+c)=p^3-3pq$$
Thank you
Asked
Active
Viewed 1,564 times
1 Answers
7
Note that: $$ (a^3+b^3+c^3)\geq a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ac) $$ Therefore it is enough to prove that $$ (a+b+c)(ab+bc+ac)\ge 6abc $$ This is done as follows: $$ (a+b+c)(ab+bc+ac)\ge 3(abc)^{\frac{1}{3}}\times 3(abc)^{\frac{2}{3}}=9abc \geq 6abc. $$
Arash
- 11,131
-
Nice.if $a,b,c$ are real numbers,then my inequality is true? Thank you – math110 Sep 19 '13 at 11:43
-
,Thank you, I fonud maybe is true,and Now if $a,b,c$ are real numbers,How prove it? – math110 Sep 19 '13 at 11:52
-
But mean that don't use $AM-GM$ – math110 Sep 19 '13 at 12:02
-
$a=b=c=-1$, LHS$=-9$,RHS$=0$,so it is not true for all real. – chenbai Sep 19 '13 at 12:59
-
Since it is an inequality between two fifth-degree homogeneous polynomials, if it holds for a certain triple $(a,b,c)$ then it holds in the opposite direction for $(-a,-b,-c)$! – Jack D'Aurizio Sep 19 '13 at 13:04