Prove the inequality $${1 \over a^3 (b + c)} + {1 \over b^3 (a+c)} + {1 \over c^3 (a+b)} \ge \frac 32 $$ So, I know a proof for this, but I basically memorized it without understanding. It's $$ {(\frac 1a + \frac 1b + \frac 1c)}^2 + {(a(b+c) + b(a+c) + c(a+b))}^-1 = {ab + bc + ac \over 2} \ge {\frac 32} $$ I don't know what inequalities were applied here or how. Another proof would also be appreciated.
Prove the inequality ${1 \over a^3 (b + c)} + {1 \over b^3 (a+c)} + {1 \over c^3 (a+b)} \ge \frac 32 $ given that $abc = 1$
1
$\begingroup$
inequality
proof-writing
a.m.-g.m.-inequality
holder-inequality
-
0Try $a=b=c=2$, perhaps $abc=1$ – 2017-01-25
-
2Your inequality does not hold true if you take $a,b,c$ all large enough. Perhaps there are conditions you missed mentioning. – 2017-01-25
-
0i think it must be $$abc=1$$ – 2017-01-25
-
0this is a IMO sum i think – 2017-01-25
-
0Yes, $abc = 1$. I'll edit the post. – 2017-01-25
2 Answers
1
we have $$\frac{1}{a^3(b+c)}=\frac{bc}{a^2(b+c)}=\frac{(bc)^2}{a(b+c)}$$ then we get $$\frac{(bc)^2}{a(b+c)}+\frac{(ac)^2}{b(a+c)}+\frac{(ab)^2}{c(a+b)}\geq \frac{(ab+bc+ac)^2}{2(ab+bc+ac)}\geq \frac{3}{2}$$ by $AM-GM$
1
By Holder and AM-GM $$\sum\limits_{cyc}\frac{1}{a^3(b+c)}=\sum_{cyc}\frac{b^3c^3}{b+c}\geq\frac{(ab+ac+bc)^3}{3\sum\limits_{cyc}(b+c)}\geq\frac{3abc(a+b+c)(ab+ac+bc)}{3\sum\limits_{cyc}(b+c)}\geq\frac{3}{2}$$