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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?

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    Are $a,b,c \in \mathbb{N}$? – 2017-02-16
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    How are you defining $\ge$? – 2017-02-16

2 Answers 2

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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

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    I don't see how this deals, for example, with the triple $(3,2,1)$ starting from the base case. – 2017-02-16
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    As written, the above only proves the inequality for $a=b=c\,$. – 2017-02-16
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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 $2a=b+c$.

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    My purpose in giving the above answer was to allow me to make comments (I don't have the clout to add these to the original answer). – 2017-02-17
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    There are two things wrong with the original answer. The first is that no base case is provided. A standard induction not only proves that a statement $S(k+1)$follows from $S(k)$, but also that $S(a)$ is true for some natural number $a$ (the base case). It then follows that $S(n)$ is true for all $n\geq a$ (but not necessarily for $n\leq a$. – 2017-02-17
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    The second is that the induction attempts the induction step simultaneously for three variables, but only one at a time is valid. If it were valid to use multiple variables then you could prove for example $a+b+c$ is divisible by 3, using the argument $0+0+0$ is divisible by 3 (the base case) and $(a+1)+(b+1)+(c+1)$ is divisible by 3 follows from $a+b+c$ is divisible 3. – 2017-02-17
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    I think the question probably needs some supporting material. The only reason I think a proof by induction might be suggested is for example if you're proving basic properties of the natural numbers from Peano's axioms and haven't yet proved $m\geq n\Rightarrow m+k\geq n+k$. In this case we'd need to know what has been proved so far and how $m\geq n$ has been defined. – 2017-02-17