<s>[INST] Volume of a cuboid whose diagonal and surface area is known The sum of length, breadth and depth of cuboid is $19$cm and its diagonal is $5\sqrt{5}$cm.
Its volume is:

a) 125
b) 236
c) 361
d) 486

Solution:
$$\ell^2 + b^2 +h^2 = 125\quad\text{ and }\quad \ell+b+h=19,$$
How can i find volume from this information?
 [/INST] Define 
$$\begin{cases}
p_1 = & \ell + b + h\\
p_2 = & \ell^2 + b^2 + h^2\\
p_3 = & \ell^3 + b^3 + h^3
\end{cases}
\quad\text{ and }\quad
\begin{cases}
s_1 = & \ell + b + h\\
s_2 = & \ell b + \ell h + b h\\
s_3 = & \ell b h
\end{cases}
$$
We know $p_1 = s_1 = 19$ and $p_2 = 125$.  
Apply $AM \ge GM$ to the three numbers $\ell, b, h$, we get an upper bound for $s_3$:
$$s_3 = \ell b h \le \left(\frac{\ell + b + h}{3}\right)^3 = \frac{p_1^3}{3} = \frac{19}{3}^3 \sim 254.037$$
This rules out choices (c) and (d).
Apply Cauchy Schwarz inequality to $( \sqrt{\ell}, \sqrt{b}, \sqrt{h} )$ and $( \sqrt{\ell}^3, \sqrt{b}^3, \sqrt{h}^3 )$, we get
$$p_2^2 \le p_1 p_3$$
Together with the Newton's identities


*

*$p_1 = s_1$,

*$p_2 = s_1 p_1  - 2 s_2$,

*$p_3 = s_1 p_2  - s_2 p_1 + 3s_3$


We obtain a lower bound for $s_3$:
$$s_3 = \frac13 \left( p_3 - s_1 p_2 + s_2 p_1 \right)
\ge \frac13 \left( \frac{p_2^2}{p_1} - p_1 p_2 + \frac{p_1^2 - p_2}{2} p_1 \right)
=   \frac13 \left( \frac{p_2^2}{p_1} + \frac{p_1^2 - 3 p_2}{2} p_1 \right)
=   \frac13 \left( \frac{125^2}{19} + \frac{19^2 - 3\cdot 125}{2} \cdot 19\right)
= \frac{4366}{19} \sim 229.790
$$
This rules out choice (a) and leaves us choice (b) $s_3 = 236$ as the only possible answer.
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