$ a,b,c\ge 0,\ \ \ a^3+b^3+c^3+abc=4 $ Prove that $a^3b+b^3c+c^3b \le 3$
Problem
Source: SRMC 2014
Tags: inequalities, inequalities proposed
arqady
02.04.2014 05:53
It's true of course, but is there a nice proof? The equality occurs also for $(a,b,c)=(\sqrt[3]3,1,0)$.
Tima
02.04.2014 07:09
ts0_9 wrote: $ a,b,c\ge 0,\ \ \ a^3+b^3+c^3+abc=4 $ Prove that $a^3b+b^3c+c^3b \le 3$ Official solution:
if $a \geq b \geq c$ or $c \geq b \geq a$, then $c(b^2-c^2)(b-a) \le 0 <=> a^3b+b^3c+c^3a \le b(a^3+c^3+abc)$. By $A_n \geq G_n $ we have
$\sqrt[4]{b^3(\frac{a^3+c^3+abc}{3})(\frac{a^3+c^3+abc}{3})(\frac{a^3+c^3+abc}{3})} \le \frac{1}{4}(b^3 +a^3+c^3+abc) = 1 => a^3b+b^3c+c^3a \le 3.$
if $a \geq c \geq b$ or $b \geq c \geq a$, then $a^3b+b^3c+c^3a \le c(a^3+b^3+abc)$
if $b \geq a \geq c$ or $c \geq a \geq b$, then $a^3b+b^3c+c^3a \le a(c^3+b^3+abc)$
sqing
03.04.2014 11:15
The following inequality is also true. Let $ a,b,c\ge 0,\ \ \ a^3+b^3+c^3+abc=4 .$ Prove that\[a^3(b+c)+b^3(c+a)+c^3(a+b) \le 6.\]
mudok
03.04.2014 15:56
sqing wrote: The following inequality is also true. Let $ a,b,c\ge 0,\ \ \ a^3+b^3+c^3+abc=4 .$ Prove that\[a^3(b+c)+b^3(c+a)+c^3(a+b) \le 6.\] it can be proved by SRMC 2014.
sqing
04.04.2014 02:44
mudok wrote: sqing wrote: The following inequality is also true. Let $ a,b,c\ge 0,\ \ \ a^3+b^3+c^3+abc=4 .$ Prove that\[a^3(b+c)+b^3(c+a)+c^3(a+b) \le 6.\] it can be proved by SRMC 2014. Yeah. Perhaps easier to prove. Thanks.
arqady
04.04.2014 07:12
ts0_9 wrote: $ a,b,c\ge 0,\ \ \ a^3+b^3+c^3+abc=4 $ Prove that $a^3b+b^3c+c^3b \le 3$ Let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$. Hence, by Rearrangement and AM-GM we obtain: $a^3b+b^3c+c^3b=a^2\cdot ab+b^2\cdot bc+c^2\cdot cb\leq x^2\cdot xy+y^2\cdot xz+z^2\cdot yz=y(x^3+xyz+z^3)=$ $=3\sqrt[3]{y^3\left(\frac{x^3+xyz+z^3}{3}\right)^3}\leq3\sqrt[3]{\left(\frac{y^3+3\cdot\frac{x^3+xyz+z^3}{3}}{4}\right)^4}=3$.
mudok
04.04.2014 18:08
arqady wrote: $a^3b+b^3c+c^3b=a^2\cdot ab+b^2\cdot bc+c^2\cdot cb\leq x^2\cdot xy+y^2\cdot xz+z^2\cdot yz=y(x^3+xyz+z^3)=$ $=3\sqrt[3]{y^3\left(\frac{x^3+xyz+z^3}{3}\right)^3}\leq3\sqrt[3]{\left(\frac{y^3+3\cdot\frac{x^3+xyz+z^3}{3}}{4}\right)^4}=3$. It is almost the official solution. The difference is: in the official solution there is no the word "rearrangement". By the way, in the official solution, it is enough to prove only for one case $(b-a)(b-c)\le 0$, because one of the numbers \[(a-b)(a-c), \ \ (b-a)(b-c), \ \ (c-a)(c-b)\] is non-positive, so, without loss of generality we may assume $(b-a)(b-c)\le 0$.