What Is Bigger $100^{100}$or $\sqrt{99^{99} \cdot 101^{101}}$

Question

Hello every what is bigger $100^{100}$or $\sqrt{99^{99} \cdot 101^{101}}$?

I tried to square up and I got $100^{200}$ or $99^{99} \cdot 101^{101}$ and I don't have an idea how to continue.

Answer 1

Taking logarithms, we see that we want to compare $f(100)$ and $\frac12(f(99)+f(101))$, where $f(x) = x\log x$. But $f(x)$ is a convex function (its second derivative $\frac1x$ is always positive), which means that the the secant line through $(99,f(99))$ and $(101,f(101))$ lies above the graph of the function. In particular, the fact that the midpoint of this secant line lies above the point $(100,f(100))$ on the graph is exactly the statement that $\frac12(f(99)+f(101)) > f(100)$, and so $\sqrt{99^{99}101^{101}} > 100^{100}$.

Answer 2

Let $f(x)=x\ln{x}$.

Thus, $f$ is a convex function.

Thus, $$x\ln{x}+y\ln{y}\geq2\cdot\frac{x+y}{2}\ln\frac{x+y}{2}$$ or $$x^xy^y\geq\left(\frac{x+y}{2}\right)^{x+y}.$$ Now, for $x=101$ and $y=99$ we obtain $$99^{99}101^{101}>100^{200}$$ or $$\sqrt{99^{99}101^{101}}>100^{100}.$$

Answer 3

It is “well-known” that $\log(u) \le u-1$ for all $u>0$, with strict inequality for $u \ne 1$ (see for example for each $x>1 , \frac{x-1}{x}\ < \ln x < x-1$).

It follows that for all real $x > 1$ $$ \log \left( \frac{x^{2x}}{(x-1)^{x-1} (x+1)^{x+1}}\right) = (x-1) \log \left( \frac{x}{x-1}\right) + (x+1) \log \left( \frac{x}{x+1}\right) \\ < (x-1) \left( \frac{x}{x-1} - 1 \right) + (x+1) \left( \frac{x}{x+1} - 1\right) = 1 - 1 = 0 $$ and therefore $$ x^x < \sqrt{(x-1)^{x-1} (x+1)^{x+1}} \, . $$