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Amy has a clump of iron with volume $8$. She melts it and make a cuboid (she might have iron left over). Her boyfriend Ben increased the length, width and height of the cuboid by $1$ to form a new cuboid. Show that the surface area of the new cuboid is at least two times of the volume of the new cuboid.


I made this problem, so I know a solution (using inequality). Any solution is accepted as long as it makes sense. There might be a geometric solution that I don't know! I will post my solution if no one posts it.

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  • $\begingroup$ The second cuboid is hollow right? Otherwise Ben has to decrease the density of iron drastically, which is tough at safe temperatures. $\endgroup$
    – user867070
    Dec 31, 2020 at 8:44
  • $\begingroup$ @WilliamBarnes Ben has extra iron, so the new cuboid is not hollow. $\endgroup$ Dec 31, 2020 at 9:08
  • $\begingroup$ "the surface is twice the volume" makes little to zero sense, to me. What do you mean? $\endgroup$
    – Raffaele
    Dec 31, 2020 at 11:21
  • $\begingroup$ @Raffaele Such as, if the original cuboid has dimensions $1,2,3$, then the dimensions of the new cuboid is $2,3,4$, so its surface area is $2(2\cdot3+3\cdot4+4\cdot2)=52$ and its volume is $2\cdot3\cdot4=24$. We have $52>2(24)$. $\endgroup$ Dec 31, 2020 at 12:30

3 Answers 3

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Let the sides be $a,b,c$, so $abc = 8$. We WTS

$$\sum 2(a+1)(b+1) \geq 2 \prod(a+1). $$

Dividing by 2 and expanding, we WTS

$$ a + b + c +3 \geq abc + 1 = 9$$

This is true because $ a + b + c \geq 3 \sqrt[3]{abc} = 6 $.


In the event that the volume is $ 0 \leq V \leq 8$, then use the fact that

$$3\sqrt[3]{V} + 3 \geq V + 1$$

If you're stuck, show your work and explain what you've tried.

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  • $\begingroup$ I did not notice expansion works, but please notice that $abc\le8$. $\endgroup$ Jan 2, 2021 at 9:15
  • $\begingroup$ @CulverKwan A) The question states "with volume 8", not "with volume at most 8". Please clarify. B) Have you tried to modify this approach to deal with $ V \leq 8$? If so, please show your work. $\endgroup$
    – Calvin Lin
    Jan 2, 2021 at 14:33
  • $\begingroup$ Maybe I will post a self answer. $\endgroup$ Jan 3, 2021 at 2:40
  • $\begingroup$ Posted self answer down below. $\endgroup$ Jan 3, 2021 at 4:24
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We let the sides be $a,b,c$, so $abc\le8$. Then we need to prove $$2\big((a+1)(b+1)+(b+1)(c+1)+(c+1)(a+1)\big)\ge2(a+1)(b+1)(c+1)$$ Which is equivalent to $$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}\ge1$$ We let $a'=\frac8{bc}$. As $abc\le8$, $a\le\frac8{bc}=a'$, so $a+1\le a'+1$, therefore $\frac1{a+1}\ge\frac1{a'+1}$. Also, we have $a'bc=\frac8{bc}\cdot bc=8$, so we can let $a'=\frac{2x}y,b=\frac{2y}z,c=\frac{2z}x$. Then, with above facts and the CS inequality, $$\begin{split}\frac1{a+1}+\frac1{b+1}+\frac1{c+1}&\ge\frac1{a'+1}+\frac1{b+1}+\frac1{c+1}\\&=\frac1{\frac{2x}y+1}+\frac1{\frac{2y}z+1}+\frac1{\frac{2z}x+1}\\&=\frac y{2x+y}+\frac z{2y+z}+\frac x{2z+x}\\&=\frac{\big(\frac y{2x+y}+\frac z{2y+z}+\frac x{2z+x}\big)(x+y+z)^2}{(x+y+z)^{2}}\\&=\frac{\big(\frac y{2x+y}+\frac z{2y+z}+\frac x{2z+x}\big)\big(y(2x+y)+z(2y+z)+x(2z+x)\big)}{(x+y+z)^{2}}\\&\ge\frac{(y+z+x)^2}{(x+y+z)^2}\\&=1\end{split}$$

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  • $\begingroup$ A slightly quicker approach is to use Jensens: $ \sum 1/ ( a+1) \geq 3/ ( \sqrt[3]{abc} + 1 ) \geq 1$ $\endgroup$
    – Calvin Lin
    Jan 3, 2021 at 6:15
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Let V and A be the area and volume of the new cuboid. Let L, W, H be the dimensions of the original. Then:

\begin{align*}V &= \left(L+1\right)\left(W+1\right)\left(H+1\right) \\ &= LWH + LW + WH + HL + L + W + H + 1 \\ &= LWH + \frac{LWH}{H} + \frac{LWH}{L} + \frac{LWH}{W} + L + W + H + 1 \\ &= LWH + LWH \left(\frac{1}{H} + \frac{1}{L} + \frac{1}{W}\right) + \left(L + W + H\right) + 1\end{align*}

Meanwhile:

\begin{align*}A &= 2\left(L+1\right)\left(W+1\right) + 2\left(W+1\right)\left(H+1\right) + 2\left(L+1\right)\left(H+1\right) \\ &= 2 \left(LW + WH + LH + 2L + 2W + 2H\right) + 6 \\ &= 2 \left(\frac{LWH}{H} + \frac{LWH}{L} + \frac{LWH}{W}\right) + 4\left(L + W + H\right) + 6 \\ &= 2 LWH \left(\frac{1}{H} + \frac{1}{L} + \frac{1}{W}\right) + 4\left(L + W + H\right) + 6\end{align*}

Let $$V_0 = LWH$$ $$Q = \left(\frac{1}{H} + \frac{1}{L} + \frac{1}{W}\right)$$ $$R = L + W + H$$ so then: $$V = V_0 + V_0 Q + R + 1 \hspace{2.54cm} A = 2V_0Q + 4R + 6 \:,$$ or $$\frac{A}{V} = \frac{V_0 \left(2Q\right) + \left(6 + 4R\right)}{V_0 \left(1+Q\right) + \left(1 + R\right)}$$

Still not totally clear to me, so looking at a few extreme cases (I haven't used the 8 or the inequality yet)...

For $Q$ very small, this means $R$ is very large, in which case the ratio limits to $A/V \approx 4$. Okay.

For $R$ very small, $Q$ must be very large, in which case the ratio limits to $A/V \approx 2$, but still greater. So far so good.

The in-between cases should be covered by the top-heavy fraction.

As a non-surprising sanity check, the special case $V_0=8$ with $L=W=H=2$, gives exactly $2$, not greater. $$\frac{A}{V} = \frac{8 \left(2 \cdot 3/2\right) + \left(6 + 4 \cdot 6\right)}{8 \left(1 + 3/2\right) + \left(1 + 6\right)} = \frac{24+6+24}{20+7} = \frac{54}{27} = 2$$

(I now see why you chose 8 for the original volume.)

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