Prove that $\min \{a^7,b^3,c^2,1\} \le abc$ If $a,b,c$ are three non-negative real numbers, prove that:
$$\min \{a^7,b^3,c^2,1\} \le abc$$
I tried several ideas with the geometric mean:
$$\min \{a^7,b^3,c^2,1\} \leq \sqrt[4]{a^7b^3c^2}$$
but $\sqrt[4]{a^7b^3c^2} \leq abc$ is not true. I tried with other means but nothing significant. I think these ideas fail because the inequality is not homognenous.
 A: Let our minimum be equal to $k$.
Thus, $$a^{42}\geq k^6,$$
$$b^{42}\geq k^{14},$$ $$c^{42}\geq k^{21}$$ and $$1\geq k.$$
Thus, $$(abc)^{42}\geq k^{6+14+21+1}$$ or
$$abc\geq k.$$
A: For the sake of contradiction, assume that:
$$\min\{a^7,b^3,c^2,1\} > abc$$
Then:
$$a^7 > abc \Rightarrow a > (abc)^{\frac{1}{7}}$$
$$b^3 > abc \Rightarrow b > (abc)^{\frac{1}{3}}$$
$$c^2 > abc \Rightarrow c > (abc)^{\frac{1}{2}}$$
Also, because $\frac{1}{2}+\frac{1}{3}+\frac{1}{7} < 1$, we have:
$$1 > abc\Rightarrow 1 > abc^{1-\frac{1}{2}-\frac{1}{3}-\frac{1}{7}}$$
Multiplying these four inequalities, it follows that:
$$abc > (abc)^{\frac{1}{7}} \cdot (abc)^{\frac{1}{3}}\cdot  (abc)^{\frac{1}{2}} \cdot abc^{1-\frac{1}{2}-\frac{1}{3}-\frac{1}{7}} = abc$$
a contradiction.
A: One can also grind through to get an answer:
If $a,b,c \ge 1$ or any of $a,b,c$ is zero then clearly the statement is true.
So we can suppose all are strictly positive and $\min (a,b,c) <1$.
If $a^7$ is the $\min$  then $a^{7 \over 5} \le b$ and $a^{7 \over 2} \le c$ so
$a^7 = a a^{7 \over 5} a^{7 \over 2} a^{11 \over 10} \le a a^{7 \over 5} a^{7 \over 2} \le abc$.
The same analysis mutatis mutandis applies to $b,c$.
A: For a positive $a$ and $x>1$,
$$
min(a^x,1) \leq min(a,1) \leq a 
$$
Moreover, 
$$
[min(a,b,1)]^2 = min(a^2,b^2,1) \leq min(a,1) \, min(b,1) \, .
$$
This result is immediately extended to the case of three or more variables. Apply the two equations above to obtain
$$
min(a^n,b^n,...,h^n,1) \leq a b ... h 
$$
for $n>0$ variables $a...h$. The first equation tells you that
$$
min(a^{n_1},b^{n_2},...,h^{n_n},1) \leq a b ... h 
$$
when $n_1...n_n\geq n$. 
