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.

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.
