AM-GM inequality.

Question

By the AM-GM Inequality, we know that $$(2^{1-x} + 2^{x+1} + 3^{x} +3^{-x})/4\geq(2^{1-x}\cdot2^{x+1}\cdot3^{x} \cdot3^{-x})^{1/4}$$

By solving this, we get $$(2^{1-x} + 2^{x+1} + 3^{x} +3^{-x})/4\geq\sqrt{2} \\ \implies f(x)\geq2\sqrt{2}$$

However, if AM-GM is applied separately, we get $$(2^{1-x} + 2^{x+1})/2\geq[2^{1-x}\cdot2^{x+1}]^{1/2} \\ (2^{1-x} + 2^{x+1})\geq4, \\ ( 3^{x} +3^{-x})/2\geq(3^{x}\cdot3^{-x})^{1/2}\\ 3^{x} +3^{-x}\geq2\\ \implies f(x)\geq3$$

Why are the results different?

Answer 1

In the first inequality, we have $$\frac{2^{1-x} + 2^{x+1} + 3^{x} +3^{-x}}{4}\geq\sqrt{2}\implies f(x)\geq2\sqrt{2}$$

However, its equivalent condition is:

$2^{1-x} = 2^{x+1} = 3^{x} = 3^{-x}$, which has no solutions, so $f(x)>2\sqrt2$.

(By $3^x=3^{-x}$, we have $x=0$, but $2^{0+1}\neq3^0$)

In the second one, the equivalent condition is:

$2^{1-x} = 2^{x+1}, 3^{x} = 3^{-x}$, which implies $x=0$.

So $f(x)\geq f(0)=3$, then we have the minimum of $f(x)$ is $3$.

In this case, $f(x)\geq f(0)=3>2\sqrt{2}$, so the first inequality still holds.