Author Topic: Inequality Solution (Read 624 times) Tweet Share

frog0101 · « **on:** October 12, 2018, 11:43:31 am »

Hi,
When proving an inequality, is it acceptable if we prove one like this (or no??):
Prove:
$ab+cd\leq {(a^2+c^2)^{\frac{1}{2}}(b^2+d^2)}^{\frac{1}{2}}$

$\therefore (ab+cd)^2\leq (a^2+c^2)(b^2+d^2)$
$a^2b^2+2abcd+c^2d^2\leq a^2b^2+a^2d^2+c^2b^2+c^2d^2$
$2abcd\leq a^2d^2+c^2b^2$

I wasn't sure if we are then able to finish the proof by doing this,

If the above is true, then the inequality must be true. Consider:
$(ac-cd)^2\geq 0$
$a^2d^2+c^2d^2\geq 2abcd$
$\therefore ab+cd\leq (a^2+c^2)^\frac{1}{2}(b^2+d^2)^{\frac{1}{2}}$

To me, this way of proving the inequality (and any inequality) appears the same as beginning the proof with the second last line (once you have worked out how to start, which is the first section essentially) and then working backwards (however, this new method has much, much... less writing).
So is this ok to do in the HSC (and not lose any marks for it)- expand and simplify the inequality until you have found one statement that underpins the original, and then prove that statement, and you have proved the inequality?

Thanks

frog1944 · « **Reply #1 on:** October 12, 2018, 01:52:17 pm »

The way you've currently worded it is has basically shown $ A \implies B \implies C $ so you've then stated that $ C \implies A $. If you wanted to do it that way by starting at something you're required to prove and working to some conclusion that is correct you need to show that it's an if and only if e.g $A \iff B $ means $A \implies B $ and $B \implies A $.

RuiAce · « **Reply #2 on:** October 12, 2018, 02:33:05 pm »

To overcome what's commonly referred to as the "converse fallacy", you should write the exact same backwards. So something like this.
$\text{We know that }(ac-bd)^2 \geq0.\text{ Then}\\ \begin{align*}a^2c^2 - 2abcd + b^2d^2 &\geq 0\\ 2abcd &\leq a^2d^2+c^2b^2\end{align*}$
$\text{Adding }a^2b^2+c^2d^2\text{ to both sides,}\\ \begin{align*}a^2b^2+2abcd + c^2d^2 &\leq a^2b^2+c^2d^2+a^2d^2+c^2b^2\\ (ab+cd)^2 &\leq (a^2+c^2)(b^2+d^2) \end{align*}$
$\text{Therefore taking positive square roots, since all terms are positive,}\\ ab+cd \leq (a^2+c^2)^{1/2} (b^2+d^2)^{1/2}$
With all that scrap working backwards you did, you can either
1) do that on the question booklet
2) just cross it out after you're done

Working backwards is a tool mathematicians use to help create the proof in the first place. However when actually presented, it needs to be presented in a clear, logical manner (which includes writing it only in the forwards direction).

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Author Topic: Inequality Solution (Read 624 times) Tweet Share

frog0101

Inequality Solution

frog1944

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RuiAce

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