Prove that for every real positive number $a, b, c$ with $1 \le a, b, c \le 8$ the inequality $$\frac{a+b+c}{5}\le \sqrt[3]{abc}$$
Problem
Source: INAMO Shortlist 2014 A4
Tags: algebra, inequalities, three variable inequality
r1chard_castle
22.05.2019 08:08
Let $f(a,b,c)=\sqrt[3]{abc}-\frac{a+b+c}{5}$. Since is $f(a,b,c)$ is concave for each variable, which means that it reaches minimum value when $a,b,c$ gets critical values. $f_{min}(a,b,c)=\min\{f(8,8,8),f(8,8,1),f(8,1,1),f(1,1,1)\}=0$
sqing
22.05.2019 09:12
parmenides51 wrote: Prove that for every real positive number $a, b, c$ with $1 \le a, b, c \le 8$ the inequality $$\frac{a+b+c}{5}\le \sqrt[3]{abc}$$ Proof of Zhangyunhua:
Attachments:
Nguyenhuyen_AG
22.05.2019 10:08
parmenides51 wrote: Prove that for every real positive number $a, b, c$ with $1 \le a, b, c \le 8$ the inequality $$\frac{a+b+c}{5}\le \sqrt[3]{abc}$$ Suppose $a \geqslant b \geqslant c,$ then \[125abc - (a+b+c)^3 = \left [ (4c+4a+b)(a-b)+\frac{3c(94a+277c)}{7} \right ](b-c)+\left [\frac{c(51b+47c)}{7}+(a+8b+7c)(a-b)+8(b-c)^2 \right ](8c-a).\]Therefore \[(a+b+c)^3 \leqslant 125abc.\]Equality occur when $a=8b=8c.$
luofangxiang
22.05.2019 13:54
parmenides51 wrote: Prove that for every real positive number $a, b, c$ with $1 \le a, b, c \le 8$ the inequality $$\frac{a+b+c}{5}\le \sqrt[3]{abc}$$
sqing
22.05.2019 16:53
luofangxiang wrote:
parmenides51 wrote:
Prove that for every real positive number $a, b, c$ with $1 \le a, b, c \le 8$ the inequality
$$\frac{a+b+c}{5}\le \sqrt[3]{abc}$$
Let $a, b, c\in[1,2] .$ Prove that $$a^3+b^3+c^3 \leq 5abc.$$Solution of unrulykid3: Assume that $a\ge b\ge c$. We have $$(a-2)(a^2+2a-1)\le 0 \Leftrightarrow a^3+2\le 5a, (b-1)(b^2+b+1-5a)\le 0\Leftrightarrow 5a+b^3\le 5ab+1, (c-1)(c^2+c+1-5ab)\le0\Leftrightarrow5ab+c^3\le 5abc+1,$$Adds 3 inequalities, we are done. Equality occur when $a=2b=2c.$ https://artofproblemsolving.com/community/c6h588217p3482275 https://artofproblemsolving.com/community/c6h1380833p7655148 https://artofproblemsolving.com/community/c6h1382023p7666385 https://artofproblemsolving.com/community/c260h1700387p11020960 here here
ytChen
22.05.2019 23:50
Prove that $\frac{a+b+c+d}{28}\le \sqrt[4]{abcd}$ for any real numbers $a, b, c,d$ with $1 \le a, b, c,d \le 81$.
Solution. Put $a=x^4$, $b=y^4$, $c=z^4$, $d=u^4$, where $x,y,z,u\in[1,3]$. Thus,
$$\frac{a+b+c+d}{28}\le \sqrt[4]{abcd}\iff x^4+y^4+z^4+u^4\le28xyzu.$$Without loss of generality, assume $x\ge y\ge z\ge u$. Thus, the following four inequalities hold,
\begin{align*}&(1)\quad x^4+3\le 28x\iff x^4+3-28x=(x-3)[x(x^2+3x+9)-1]\le0\\
&(2)\quad y^4-1\le 28xy-28x\iff (y-1)[(y^2+1)(y+1)-28x]\le0\\
&(3)\quad z^4-1\le 28xyz-28xy\iff (z-1)[(z^2+1)(z+1)-28xy]\le0\\
&(4)\quad u^4-1\le 28xyzu-28xyz\iff (u-1)[(u^2+1)(u+1)-28xyz]\le0.
\end{align*}(1)+(2)+(3)+(4) yields $x^4+y^4+z^4+u^4\le28xyzu$, where the equality occurs when $(x,y,z,u)=(3,1,1,1)$. As stated. $\blacksquare$
ytChen
23.05.2019 02:31
I think the proposition below is also true: Let $n\ge3$ be an integer, and $a_i$ real numbers such that $1 \le a_i \le(n-1)^n$ for every $i\in\{1,2,\cdots, n\}$. Prove that $$\frac{\displaystyle\sum_{i=1}^na_i }{(n-1)^{n-1}+1}\le\prod_{i=1}^n\sqrt[n]{a_i}.$$
sqing
23.05.2019 03:04
ytChen wrote:
Prove that $\frac{a+b+c+d}{28}\le \sqrt[4]{abcd}$ for any real numbers $a, b, c,d$ with $1 \le a, b, c,d \le 81$.
ProofSolution. Put $a=x^4$, $b=y^4$, $c=z^4$, $d=u^4$, where $x,y,z,u\in[1,3]$. Thus,
$$\frac{a+b+c+d}{28}\le \sqrt[4]{abcd}\iff x^4+y^4+z^4+u^4\le28xyzu.$$Without loss of generality, assume $x\ge y\ge z\ge u$. Thus, the following four inequalities hold,
\begin{align*}&(1)\quad x^4+3\le 28x\iff x^4+3-28x=(x-3)[x(x^2+3x+9)-1]\le0\\
&(2)\quad y^4-1\le 28xy-28x\iff (y-1)[(y^2+1)(y+1)-28x]\le0\\
&(3)\quad z^4-1\le 28xyz-28xy\iff (z-1)[(z^2+1)(z+1)-28xy]\le0\\
&(4)\quad u^4-1\le 28xyzu-28xyz\iff (u-1)[(u^2+1)(u+1)-28xyz]\le0.
\end{align*}(1)+(2)+(3)+(4) yields $x^4+y^4+z^4+u^4\le28xyzu$, where the equality occurs when $(x,y,z,u)=(3,1,1,1)$. As stated. $\blacksquare$
ytChen wrote:
I think the proposition below is also true:
Let $n\ge3$ be an integer, and $a_i$ real numbers such that $1 \le a_i \le(n-1)^n$ for every $i\in\{1,2,\cdots, n\}$. Prove that
$$\frac{\displaystyle\sum_{i=1}^na_i }{(n-1)^{n-1}+1}\le\prod_{i=1}^n\sqrt[n]{a_i}.$$