Let G be a planar graph with at least 3 verticies. Prove that G contains at least 3 verticies whose degree is $\leq 5$.
What i have tried to do:
Lets suposse that there exist a planar graph with at least 3 verticies that has only at most 2 verticies whose degree is $\leq 5$. Therafore the sum of verticies degress in that graph :
$2E \geq 6(n-2) \implies E \geq 3n - 6$
from the Newton i know that
$E \leq 3n - 6$
So $E=3n-6$ and there is no contradiction so prove fails
any ideas how to prove it
I found it out, forgot that i can use Euler only when it is connected
Let G be the graph with the smallest verticies number but greater or equal 3 who has at most 2 verticies with degree lesser or equal 5. G must be connected graph cuz if wasn't i could take connected subgraph out of him who would still be counterexample so G wouldn't be smallest).
Sum of verticies degrees in graph must satisfy this inequality:
2E >= 6(n-2) + 2 || 6(n-2) becouse only 2 verticies are allowed to be have lesser degree then 6 and + 2 becouse G is connected so there are no verticies with degree 0 (so reamaning 2 verticies have at least degree equal 1.
After some math:
E>= 3n - 5 from equler E <= 3n - 6
Got contradiction. So G must have more then 2 verticies with degree less or equal 5