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Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b)(a+c)(b+c)=8$. Prove that: $$\frac{a+b+c}{3}\geq\sqrt[53]{\frac{a^4+b^4+c^4}{3}}$$ I think $uvw$ does not help here.

My another similar inequality is very easy:

with the same condition prove that: $$\frac{a+b+c}{3}\geq\sqrt[27]{\frac{a^3+b^3+c^3}{3}}$$ My proof:

By AM-GM $(a+b+c)^3=a^3+b^3+c^3+24\geq9\sqrt[9]{(a^3+b^3+c^3)\cdot3^8}$

and we are done!

Thank you!

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  • $\begingroup$ What is $u v w$? $\endgroup$ Feb 22 '18 at 0:11
  • 1
    $\begingroup$ See here artofproblemsolving.com/community/c6h278791 $\endgroup$ Feb 22 '18 at 5:02
  • $\begingroup$ Where does this inequality come from? $\endgroup$ Feb 25 '18 at 10:18
  • $\begingroup$ @Wei-Cheng Liu It's mine. $\endgroup$ Feb 25 '18 at 11:14
  • $\begingroup$ Continuing with the inequality that you already proved, it suffices to show $ \sqrt[27]{\frac{a^3+b^3+c^3}{3}} \geq \sqrt[53]{\frac{a^4+b^4+c^4}{3}} $ and, after using the extra condition, a 2-d plot shows that this actually holds true. $\endgroup$
    – Andreas
    Mar 17 '18 at 18:37
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(Homogenization)We need to prove that (for $a, b, c \ge 0$, at most one of them is zero) \begin{align} \frac{a+b+c}{3} \ge \sqrt[53]{\frac{a^4+b^4+c^4}{3}\Big(\frac{(a+b)(b+c)(c+a)}{8}\Big)^{49/3}}. \end{align} WLOG, assume that $c=1$ and $a+b \ge 1$. Let $p=a+b, \ q = ab$. Using $a^4+b^4 = p^4-4p^2q+2q^2$, taking logarithm on both sides, it suffices to prove that \begin{align} f(p, q) = 53\ln \frac{p+1}{3} - \ln \frac{p^4-4p^2q+2q^2+1}{3} - \frac{49}{3}\ln p - \frac{49}{3}\ln (1+p+q) + \frac{49}{3}\ln 8 \ge 0. \end{align} Since $q \le \frac{p^2}{4}$ and $p\ge 1$, we have $52p^2-3p-3-55q \ge 52p^2-3p-3-55(\frac{p^2}{4}) = \frac{153}{4}p^2-3p-3 > 0$. Thus, we have \begin{align} \frac{\partial f}{\partial q} &= -\frac{2(52p^2-3p-3-55q)^2 - 2713\, p^4 - 36\, p^3 - 54\, p^2 - 36\, p + 2677}{165(p^4-4p^2q+2q^2+1)(1+p+q)}\\ &\le -\frac{2(\frac{153}{4}p^2-3p-3)^2 - 2713\, p^4 - 36\, p^3 - 54\, p^2 - 36\, p + 2677}{165(p^4-4p^2q+2q^2+1)(1+p+q)}\\ &= -\frac{31p^4-72p^3-72p^2+392}{24(p^4-4p^2q+2q^2+1)(1+p+q)}\\ &\le 0. \end{align} Thus, we have $f(p, q) \ge f(p, \frac{p^2}{4})$. It suffices to prove that $f(p, \frac{p^2}{4})\ge 0$ or $$g(p) = 53\ln \frac{p+1}{3} - \ln \frac{p^4+8}{24} - \frac{49}{3}\ln p - \frac{49}{3}\ln (1+p+\frac{p^2}{4}) + \frac{49}{3}\ln 8\ge 0.$$ Note that $$g'(p) = \frac{(p-2)(37\, p^4 - 48\, p^3 - 96\, p^2 - 96\, p + 392)}{3(2+p)(p+1)(p^4+8)p}.$$ Since $37\, p^4 - 48\, p^3 - 96\, p^2 - 96\, p + 392 > 0$, we have $g'(p) < 0$ for $1\le p < 2$ and $g'(p) > 0$ for $2 < p$. Thus, $g(p) \ge g(2) = 0.$ We are done.

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  • $\begingroup$ Thank you very much! But this idea does not work always. It's better to say that it hardly ever works. $\endgroup$ Jun 21 '19 at 12:55
  • $\begingroup$ I agree. By the way, the KKT conditions (or Lagrange multiplier) can also be used easily to get the result that two of $a, b, c$ are equal. $\endgroup$
    – River Li
    Jun 21 '19 at 13:00
  • $\begingroup$ Can you show? I tried to use LM, but without success. The system turns out very ugly. Also, why for equality case of two variables we'll get unique minimum points? $\endgroup$ Jun 21 '19 at 13:07
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    $\begingroup$ If you prove that two of them are equal using any method, WLOG, assume that $a=b$, you may obtain $c = \frac{2-a\sqrt{a}}{\sqrt{a}}$ from $(a+b)(b+c)(c+a)=8$. Let $a = u^2$ and $c = \frac{2-u^3}{u}$. Then prove $f(u) = \ln \frac{a+b+c}{3} - \frac{1}{53}\ln \frac{a^4+b^4+c^4}{3} \ge 0.$ $\endgroup$
    – River Li
    Jun 21 '19 at 13:58
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Another solution using KKT conditions (or Lagrange multiplier)

Let \begin{align} L = \ln \frac{a+b+c}{3} - \frac{1}{53}\ln \frac{a^4+b^4+c^4}{3} - t ((a+b)(b+c)(c+a) - 8). \end{align} From $\frac{\partial L}{\partial a} = \frac{\partial L}{\partial b} = \frac{\partial L}{\partial c} = 0$, we have \begin{align} 0 = \frac{\partial L}{\partial a} - \frac{\partial L}{\partial b}= \frac{\left(a - b\right)\, \left((53a^5+53a^4b+53ab^4+53ac^4+53b^5+53bc^4)t-4a^2-4ab-4b^2\right)}{53(a^4 + b^4 + c^4)},\quad \cdots\cdots\cdots (2)\\ 0 = \frac{\partial L}{\partial b} - \frac{\partial L}{\partial c} = \frac{\left(b - c\right)\, \left((53a^4b+53a^4c+53b^5+53b^4c+53bc^4+53c^5)t-4b^2-4bc-4c^2\right)}{53(a^4 + b^4 + c^4)}. \quad \cdots\cdots\cdots(3) \end{align} We claim that two of $a, b, c$ are equal. Otherwise, from (2), we have $$t= \frac{4(a^2+ab+b^2)}{53(a^5 + a^4\, b + a\, b^4 + a\, c^4 + b^5 + b\, c^4)}$$ which, when inserted into (3), results in $$\frac{4}{53}\frac{\left(a - c\right)\, \left(b-c\right)\, \left(a\, b + a\, c + b\, c\right)}{\left(a + b\right)\, \left(a^4 + b^4 + c^4\right)} =0.$$ Contradiction.

Further we claim that $a = b = c$. Otherwise, WLOG, assume that $a = b \ne c$, from (3), we have $$t = \frac{4}{53}\frac{a^2 + a\, c + c^2}{2\, a^5 + 2\, a^4\, c + a\, c^4 + c^5}$$ which results in $$0 = \frac{\partial L}{\partial a} = -\frac{1}{53}\frac{ - 74\, a^4 + 48\, a^3\, c + 48\, a^2\, c^2 + 24\, a\, c^3 - 49\, c^4}{\left(2\, a^4 + c^4\right)\, \left(2\, a + c\right)}.$$ However, $- 74\, a^4 + 48\, a^3\, c + 48\, a^2\, c^2 + 24\, a\, c^3 - 49\, c^4\ne 0$ which follows from $74x^4-48x^3-48x^2-24x+49 > 0$ for all real numbers $x$. Contradiction. We are done.

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