I will present a slight generalization: For any set of numbers $a_i$ such that $\sum_{cyc}a_i^{p_1} = \sum_{cyc} a_i^{p_2}$ where $p_1 > p_2$, then $\prod_{cyc}a_i^{a_i^{p_1}} \ge 1$. We can rearrange the initial condition to get: \[\frac{\displaystyle\sum_{cyc} a_i^{p_2}}{\displaystyle\sum_{cyc} a_i^{p_1}} = 1\] By weighted AM-GM, we now have: \begin{align*} 1 &= 1^{\frac{\displaystyle\sum_{cyc} a_i^{p_1}}{p_1 - p_2}} \\ &= \frac{\displaystyle\sum_{cyc} a_i^{p_2}}{\displaystyle\sum_{cyc} a_i^{p_1}}^{\frac{\displaystyle\sum_{cyc} a_i^{p_1}}{p_1 - p_2}} \\ &= \frac{\displaystyle\sum_{cyc} a_i^{p_1}\frac{1}{a_i^{p_1 - p_2}}}{\displaystyle\sum_{cyc} a_i^{p_1}}^{\frac{\displaystyle\sum_{cyc} a_i^{p_1}}{p_1 - p_2}} \\ &\ge \left(\left(\displaystyle\prod_{cyc}\left(\frac{1}{a_i^{p_1 - p_2}}\right)^{a^{p_1}}\right)^{\frac{1}{\displaystyle\sum_{cyc} a_i^{p_1}}}\right)^{\frac{\displaystyle\sum_{cyc} a_i^{p_1}}{p_1 - p_2}}\\ &= \displaystyle\prod_{cyc}a_i^{a_i^{-p_1}} \end{align*} Dividing both sides by $\prod_{cyc}a_i^{a_i^{-p_1}}$ yields $\prod_{cyc}a_i^{a_i^{p_1}} \ge 1$, as desired. Letting $p_1 = 1$, $p_2 = \frac{1}{7}$, and setting t he number of elements in $a_i$ to $3$ gives us the solution to the question.

One-liner: By weighted GM-HM, \[\sqrt[\frac76a+\frac76b+\frac76c]{(a^{6/7})^{\frac76a}(b^{6/7})^{\frac76b}(c^{6/7})^{\frac76c}}\ge\left(\frac{\frac76a(a^{6/7})^{-1}+\frac76b(b^{6/7})^{-1}+\frac76c(c^{6/7})^{-1}}{\frac76a+\frac76b+\frac76c}\right)^{-1}\] And the RHS is clearly 1 by the given condition.

We show that \[\frac{6}{7} \ln {x} \ge 1-x^{-\frac{6}{7}}\]. Consider the function \[f(x) = \frac{6}{7} \ln {x} - 1-x^{-\frac{6}{7}}\] Taking a derivative shows that it achieves a minimum at $\frac{6}{7x} = \frac{6}{7}x^{-\frac{13}{7}} \implies x = 1$ or $x = 0$ is a critical point. But at $x = 1$ or $x$ approaches $0$, the inequality holds, and we see that the LHS is at least the RHS as x approaches infinity. So our critical point was the minimum. Note that at $x = 1$, the LHS equals the RHS. So the inequality is proven. Now \[\sum{a \cdot (\frac{6}{7}\ln{a})} \ge \sum{a-a^{\frac{1}{7}}} = 0 \implies \] \[\sum{a \ln{a}} \ge 0 \implies \]\[a^ab^bc^c \ge 1\].

^ You can actually prove the first fact by simply using $e^t \ge 1+t$ with $t=\ln x^{-\frac{6}{7}}$. The official solution is similar to the GM-HM above. Using weighted AM-GM, \[ 1 = \sum_{\text{cyc}} \frac{a}{a+b+c} \cdot a^{-\frac{6}{7}} \ge (a^ab^bc^c)^{-\frac{6/7}{a+b+c}} \] whence $a^ab^bc^c\ge1$.

Can you also do this by showing that if a^1/7+b^1/7= a+b, then a^a *b^b>1? I mean if a^1/7+b^1/7= a+b then you can write the same for a and c or b and c and add it, then it's true for 3 variables too. You can multiply the products to get the desired result. Are my missing something?

But you can't assume that $a^{1/7}+b^{1/7}=a+b$...

Another similar solution using GM-HM is $a^{\frac{a}{a+b+c}}b^{\frac{b}{a+b+c}}c^{\frac{c}{a+b+c}} \ge \frac{3}{\frac{a}{a+b+c}\frac{1}{a}+\frac{b}{a+b+c}\frac{1}{b}+\frac{c}{a+b+c}\frac{1}{c}} = a+b+c$. And it we had $a+b+c < 1$ then we would have $a, b, c < 1 \rightarrow a^{\frac{1}{7}} > a, b^{\frac{1}{7}} > b, c^{\frac{1}{7}} > c$ which would contradict the problem's given statement. Taking each side of the inequality to the power $\frac{1}{a+b+c}$ yields the desired result. Note that my method also solves the generalization above, because if we have $\displaystyle\sum_{i = 1}^n a_i^{p_1} = \displaystyle\sum_{i = 1}^n a_i^{p_2}$ with $p_1 > p_2$ then setting $b_i = a_i^{p_1}$, we have $\displaystyle\sum_{i = 1}^n b_i = \displaystyle\sum_{i = 1}^n b_i^{\frac{p_2}{p_1}}$. Also, raising the desired inequality to the power of $p_1$ yields $\prod_{i = 1}^n b_i^{b_i} \ge 1$, which is easily proven by the same argument given that $\frac{p_2}{p_1} < 1$.

So this was my solution on the test: Change it to $\sum x^7 = \sum x$ and we want to show $\sum x^7 \text{ln} x \geq 0$. Note by holder's \[ (\sum x^7)^5 ( \sum x ) \geq ( \sum x^6 )^6 \implies \sum x^7 \geq \sum x^6 \] Now by Jensen's on the convex function $x \text{ln} x$ and the weights $\frac{x^6}{x^6+y^6+z^6}$ , we get that \[ \frac{1}{\sum x^6} \sum x^7 \text{ln} x \geq \frac{\sum x^7}{\sum x^6} \text{ln} ( \frac{\sum x^7}{\sum x^6} ) \geq 0 \Box \]

mathuz wrote: $xlnx$ - convex? I don't understand. $(x\ln x)''=\frac{1}{x}>0$.

MathPanda1 wrote: v_Enhance wrote: The official solution is similar to the GM-HM above. Using weighted AM-GM, \[1 = \sum_{\text{cyc}} \frac{a}{a+b+c} \cdot a^{-\frac{6}{7}} \ge (a^ab^bc^c)^{-\frac{6/7}{a+b+c}} \] whence $a^ab^bc^c\ge1$. Hi v_Enhance, what is the motivation for this very beautiful solution? Thank you very much! Well, I wrote the problem starting from the line \[ 1 = \sum_{\text{cyc}} \frac{a}{a+b+c} a^s \ge \prod_{\text{cyc}} a^{s \frac{a}{a+b+c}} \] i.e. I wanted to use Weighted AM-GM in such a way that one obtained exponents $a$, $b$, $c$. Plugging in $s = -\tfrac67$ yields the problem. Maybe you can think of it that way.

nprime06 wrote:

one nice solution

How do I express something like the cube root of a number? please give me the latex

Use $\sqrt[3]{x}$

Let $x^7 = a , y^7 = b , z^7 = c$ We want

Putting it back in gives $$\sum{x^7 ln{x}} \ge \sum \frac{x^7}{6} - \frac{x}{6} = 0$$