Proof inequality (1 Viewer)
- Thread starter D3spair
- Start date
Have you learnt about the arithmetic/geometric mean inequality yet? It usually looks something like 
. This can be extended to 
, 
etc.
There's a shortcut to the question using this method:
^3 \geq \left (\sqrt[3]{abc}\right )^3 \rightarrow \frac{(a+b+c)^3}{27} \geq {abc} \rightarrow (a+b+c)^3 \geq {27abc} )
However, since the question asks you to expand, you can collect the 'like' terms of
, 
and 
, and do a similar process. You should end up with 
.
Hopefully you find this helpful
There's a shortcut to the question using this method:
However, since the question asks you to expand, you can collect the 'like' terms of
Hopefully you find this helpful
yup I’ve learned am-gm etc., just needed assistance on what to do after expanding. ThanksHave you learnt about the arithmetic/geometric mean inequality yet? It usually looks something like
There's a shortcut to the question using this method:
However, since the question asks you to expand, you can collect the 'like' terms of
Hopefully you find this helpful
You have that ^3 = x^3 + y^3 + z^3 + 3(x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2) + 6xyz)
We have three groups of terms, the last of which is already in the desired form, so we need to consider the other two groups.
First, the terms in the form
:
Expanding^2 \ge 0)
gives 
, which is the 
case of the AM-GM inequality
Put
and 
gives 
Make similar substitutions with
and 
to get 
, and then to get 
, and you have shown that:
)
Next, we need the
case of AM-GM, 
:
Taking then
, 
, and 
etc gives
It follows that we now have
 \ge 6xyz)
The case of AM-GM also deals with our cubic terms:
Taking
, 
, and 
gives 
and thus
^3 &= x^3 + y^3 + z^3 + 3(x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2) + 6xyz \\ &\ge 3xyz + 3(6xyz) + 6xyz \\ (x + y + z)^3 &\ge 27xyz \end{align*})
completing the proof. Note, however, that while I have proved the case of AM-GM, I have used the 
case without proof, and this is not appropriate unless you are given it as a result. I recommend knowing a proof of AM-GM by heart for the cases , , and 
. Of these, it is the case that is difficult, IMO...
We have three groups of terms, the last of which is already in the desired form, so we need to consider the other two groups.
First, the terms in the form
Expanding
Put
Make similar substitutions with
Next, we need the
Taking then
It follows that we now have
The
Taking
completing the proof. Note, however, that while I have proved the
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