proof $2a>b+c$ through induction

Question

assume that we have following:

$$ a \geq b; a \geq c; a,b,c \in \mathbb N $$

now proof through induction that

$$ 2a \geq b+c $$

well for $a=b=c=0$ is the above eq. true.

how would you solve this?

Answer 1

We just need to prove it for the triple $(a+1, b+1, c+1)$ assuming it's true for $(a,b,c)$.

So we want to show $$ 2(a+1) \geq (b+1) + (c+1)$$ which is the same as $$ 2a + 2 \geq (b+c) + 2$$ which is true iff $2a \geq b+ c$ which is what we have assumed

EDIT: all assuming $a,b,c \in \mathbb{N}$ 2nd EDIT: As mentioned, this only proves for $a =b =c$, however the other cases are similar

Answer 2

If you accept the first answer then you can prove more! Simply reversing the $\geq$ sign in the answer gives you $2a\leq b+c$ by exactly the same argument.

So in fact $2a=b+c$.