An algebra problem by Viki Zeta

Consider the quotient, $Q = \dfrac {2^{100}+3^{100}}{4^{100}}$ . If $Q > 1$ , then $2^{100}+3^{100} > 4^{100}$ . If $Q < 1$ , then $2^{100}+3^{100} < 4^{100}$ .

$\begin{aligned} Q & = \frac {2^{100}+3^{100}}{4^{100}} = \left(\frac 24\right)^{100} + \left(\frac 34\right)^{100} = \left(\frac 12\right)^{100} + \left(\frac 34\right)^{100} \end{aligned}$

Since $\left(\dfrac 12\right)^{100} < \left(\dfrac 12\right)^2$ and $\left(\dfrac 34\right)^{100} < \left(\dfrac 34\right)^2$ , then:

$\begin{aligned} Q & < \left(\dfrac 12\right)^2 + \left(\dfrac 34\right)^2 = 0.25 + 0.5625 = 0.8125 < 1 \end{aligned}$

Therefore, the answer is $\boxed{\text{True}}$ .

$100 < 200 \\ 2^{100} < 2^{200} \\ 2^{100} < 4^{100} \text{ ... } \fbox{ 1 } \\ 3 < 4 \\ 3^{100} < 4^{100} \text{ ... } \fbox{ 2 } \\ \text{from } \fbox{ 1 } \text{ and } \fbox{ 2 } \\ 2^{100} + 3^{100} \le 4^{100} \\ \implies 2^{100} + 3^{100} < 4^{100}$