# Prove that $\frac{a+b+c}{3}\geq\sqrt[53]{\frac{a^4+b^4+c^4}{3}}$

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!

• What is $u v w$? Feb 22 '18 at 0:11
• Feb 22 '18 at 5:02
• Where does this inequality come from? Feb 25 '18 at 10:18
• @Wei-Cheng Liu It's mine. Feb 25 '18 at 11:14
• 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. Mar 17 '18 at 18:37

## 2 Answers

(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.

• Thank you very much! But this idea does not work always. It's better to say that it hardly ever works. Jun 21 '19 at 12:55
• 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. Jun 21 '19 at 13:00
• 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? Jun 21 '19 at 13:07
• 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 $f(u) = \ln \frac{a+b+c}{3} - \frac{1}{53}\ln \frac{a^4+b^4+c^4}{3} \ge 0.$ Jun 21 '19 at 13:58

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.