Inequalities question help (1 Viewer)

devtrivedi · Oct 5, 2020

I need help with this question

fan96 · Oct 6, 2020

Hint: Make the substitutions $a = e^\alpha, b = e^\beta, c = e^\gamma$ .

Substituting, we note that $\alpha,\beta,\gamma > 0$ and the inequalities become

$\frac{a}{b} \ge \frac{c}{a} \iff e^{2\alpha} \ge e^{\beta+\gamma},$

and

$\frac{\log a}{\log b} \ge \frac{\log c}{\log a} \iff \alpha^2 \ge \beta \gamma.$

Now $e^{2\alpha} \ge e^{\beta+\gamma}$ means that

$\alpha \ge \frac{\beta + \gamma}{2},$

And $\rm{RHS} \ge \sqrt{\beta \gamma}$ by the AM-GM inequality. Squaring, we get

$\alpha^2 \ge \beta \gamma,$

completing the proof.

devtrivedi · Oct 6, 2020

thanks buddy

beetree1 · Oct 6, 2020

That was super impressive

fan96 said:
Hint: Make the substitutions $a = e^\alpha, b = e^\beta, c = e^\gamma$ .

Substituting, we note that $\alpha,\beta,\gamma > 0$ and the inequalities become

$\frac{a}{b} \ge \frac{c}{a} \iff e^{2\alpha} \ge e^{\beta+\gamma},$

and

$\frac{\log a}{\log b} \ge \frac{\log c}{\log a} \iff \alpha^2 \ge \beta \gamma.$

Now $e^{2\alpha} \ge e^{\beta+\gamma}$ means that

$\alpha \ge \frac{\beta + \gamma}{2},$

And $\rm{RHS} \ge \sqrt{\beta \gamma}$ by the AM-GM inequality. Squaring, we get

$\alpha^2 \ge \beta \gamma,$

completing the proof.

devtrivedi · Oct 6, 2020

beetree1 said:
That was super impressive

true

CM_Tutor · Oct 6, 2020

devtrivedi said:
View attachment 29085
I need help with this question

$\begin{align*} \frac{a}{b} &\ge \frac{c}{a} \\ a^2 &\ge bc \\ \ln{a^2} &\ge \ln{bc} \\ 2\ln{a} &\ge \ln{b} + \ln{c} \\ \ln{a} &\ge \frac{\ln{b}+\ln{c}}{2} \, \text{ . . . (*)} \\ \\ \text{Now, according to the AM-GM inequality,} \, \frac{x+y}{2} &\ge \sqrt{xy} \text{ for all positive real numbers $x$ and $y$} \\ \\ \text{Putting $x = \ln{b}$ and $y = \ln{c}$:} \, \frac{\ln{b}+\ln{c}}{2} &\ge \sqrt{\ln{b}\times\ln{c}} \, \text{ . . . (**)} \\ \\ \text{Combining (*) and (**):} \, \ln{a} &\ge \sqrt{\ln{b}\times\ln{c}} \\ (\ln{a})^2 &\ge \ln{b}\times\ln{c} \\ \frac{\ln{a}}{\ln{b}} &\ge \frac{\ln{c}}{\ln{a}} \end{align*}$

devtrivedi · Oct 6, 2020

CM_Tutor said:
$\begin{align*} \frac{a}{b} &\ge \frac{c}{a} \\ a^2 &\ge bc \\ \ln{a^2} &\ge \ln{bc} \\ 2\ln{a} &\ge \ln{b} + \ln{c} \\ \ln{a} &\ge \frac{\ln{b}+\ln{c}}{2} \, \text{ . . . (*)} \\ \\ \text{Now, according to the AM-GM inequality,} \, \frac{x+y}{2} &\ge \sqrt{xy} \text{ for all positive real numbers $x$ and $y$} \\ \\ \text{Putting $x = \ln{b}$ and $y = \ln{c}$:} \, \frac{\ln{b}+\ln{c}}{2} &\ge \sqrt{\ln{b}\times\ln{c}} \, \text{ . . . (**)} \\ \\ \text{Combining (*) and (**):} \, \ln{a} &\ge \sqrt{\ln{b}\times\ln{c}} \\ (\ln{a})^2 &\ge \ln{b}\times\ln{c} \\ \frac{\ln{a}}{\ln{b}} &\ge \frac{\ln{c}}{\ln{a}} \end{align*}$

thank you

Inequalities question help (1 Viewer)

devtrivedi

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