Utilizing AM $\ge$ GM in third, less obvious question

Question

In both $1st$ and $2nd$ question, I found it easy to spot where I can use the AM$\ge$GM theorem because the values are written so that it can easily be written in the form $(A+B)^2$.

However, in the $3rd$ question, I cannot find a method to combine the $2$ statements such that they go in the AM$\ge$GM form. Any hint to reach there will be preferred over an outright solution.

Answer 1

Squaring $a+b+c = 1$ gives $$ a^2+b^2 + c^2 + 2(ab+ac+bc) = 1 $$ so what remains to show is that $a^2+b^2+c^2\geq ab+ac+bc$. The proof of this follows more or less the same idea as the standard proof of AM-GM for two terms.

Answer 2

For the $c)$, we have: $3ab + 3bc + 3ca = 2ab + 2bc + 2ca + ab + bc + ca \le 2ab + 2bc + 2ca + \dfrac{a^2+b^2}{2} + \dfrac{b^2+c^2}{2} + \dfrac{c^2+a^2}{2} = 2ab + 2bc + 2ca + a^2+b^2+c^2 = (a+b+c)^2 = 1^2 = 1.$