# exponential inequality $\frac{2^{2C_1}+2^{2C_2}-1}{2^{2C_1}}\geq\frac{2^{2\hat{C_1}}+2^{2\hat{C_2}}-1}{2^{2\hat{C_1}}}$

I am trying to see if this inequality is correct or not:

## $\frac{2^{2C_1}+2^{2C_2}-1}{2^{2C_1}}\geq\frac{2^{2\hat{C_1}}+2^{2\hat{C_2}}-1}{2^{2\hat{C_1}}}$

Given that: $C_1\geq C_2$, $\hat{C_1}\geq \hat{C_2}$, $C_1\geq\hat{C_1}$,$C_2\geq\hat{C_2}$.

My attempt:

$2^{2C_1+2\hat{C_1}}+2^{2C_2+2\hat{C_1}}-2^{2\hat{C_1}}\geq 2^{2\hat{C_1}+2C_1}+2^{2\hat{C_2}+2C_1}-2^{2C_1}$

$2^{2C_2+2\hat{C_1}}-2^{2\hat{C_1}}\geq 2^{2\hat{C_2}+2C_1}-2^{2C_1}$

$2^{2C_1}+2^{2C_2+2\hat{C_1}}\geq 2^{2\hat{C_1}}+2^{2\hat{C_2}+2C_1}$

• Get rid of the 2 in front of each C and call them a b c d. It will be a lot easier to work with. – marty cohen May 30 '18 at 11:37
• @Lee Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… – gimusi Jun 22 '18 at 21:06

Consider your last line $$2^{2C_1}+2^{2C_2+2\hat{C_1}}\geq 2^{2\hat{C_1}}+2^{2\hat{C_2}+2C_1}$$ Select $C_1 = 5$, $\hat{C_1} = 2$, $C_2 = 2$,$\hat{C_2} = 1$.

This satisfies the given conditions: $C_1\geq C_2$, $\hat{C_1}\geq \hat{C_2}$, $C_1\geq\hat{C_1}$,$C_2\geq\hat{C_2}$.

Plugging this into the eq: $$2^{2\cdot 5}+2^{2\cdot 2+2 \cdot 2}\geq 2^{2\cdot 2}+2^{2\cdot 1+2 \cdot 5}$$ we see that this is violated since the numbers are $$2^{2\cdot 5}+2^{2\cdot 2+2 \cdot 2} = 1280\geq 2^{2\cdot 2}+2^{2\cdot 1+2 \cdot 5} = 4112$$ So the inequality does not hold.

This didn't materialize out of thin air, here is some more explanation how I "made this up":

With some real $a$, multiply both sides of your last line with $2^{4 a}$ which doesn't change the inequality. You get $$2^{2(C_1+2a)}+2^{2(C_2+a)+2(\hat{C_1} +a)}\geq 2^{2(\hat{C_1}+2a)}+2^{2(\hat{C_2}+a)+2(C_1+a)}$$ Shifting all variables by $a$ also doesn't change the given conditions. So in the shifted variables, you have $$2^{2a +2C_1}+2^{2C_2+2\hat{C_1}}\geq 2^{2a +2\hat{C_1}}+2^{2\hat{C_2}+2C_1}$$ which gives that
$$2^{2a}\geq \frac{2^{2\hat{C_2}+2C_1} - 2^{2C_2+2\hat{C_1}}}{2^{2{C_1}} - 2^{2\hat{C_1}}}$$ is required. Now you can choose some values on the RHS and then compute an $a$ which actually violates this.

I chose $C_1 = 4$, $\hat{C_1} = 1$, $C_2 = 1$,$\hat{C_2} = 0$ - remember these are the shifted variables. Then $$2^{2a}\geq \frac{2^{8} - 2^{4}}{2^{8} - 2^{2}} \simeq 0.95$$ or $a > -0.035$. So this can be violated by, say, $a = -1$. Then we have the original values $C_1 = 5$, $\hat{C_1} = 2$, $C_2 = 2$,$\hat{C_2} = 1$ which are used above.

• Thank you @Andreas – Lee May 31 '18 at 6:00

The inequality is not true in general, indeed we have that

$$\frac{2^{2C_1} +2^{2C_2} -1}{2^{2C_1}}\geq\frac{2^{2\hat{C_1}}+2^{2\hat{C_2}}-1}{2^{2\hat{C_1}}}\iff 1 +4^{C_2-C_1}-4^{-C_1}\ge1 +4^{\hat C_2-\hat C_1}-4^{-\hat C_1}$$ $$\iff 4^{C_2-C_1}-4^{-C_1}\ge 4^{\hat C_2-\hat C_1}-4^{-\hat C_1}$$ $$\iff 4^{-C_1}(4^{C_2}-1)\ge 4^{-\hat C_1}(4^{\hat C_2}-1)$$

now let consider

$$C_2=\hat C_2=\hat C_1>0\implies 4^{-C_1} (4^{C_2}-1)\ge 4^{-C_2}(4^{C_2}-1)\implies 4^{-C_1}\ge 4^{-C_2}$$

which is not true since $4^{-x}$ is decreasing and $C_1\ge C_2$.

• Thank you @gimusi – Lee May 31 '18 at 6:00
• @Lee You are welcome! Bye – gimusi May 31 '18 at 6:02