2
$\begingroup$

I need to prove the inequality $e^{2x-\frac{2x^2}{3}} \geq \frac{1+x}{1-x+\frac{2x^2}{3}}$ for all $x \leq 3$. This is true according to Wolfram Alpha. Any ideas?

2 Answers 2

1

We need to prove that $f(x)\geq0$, where $$f(x)=\frac{2x(3-x)}{3}-\ln(x+1)+\ln\left(1-x+\frac{2x^2}{3}\right)$$ for $-1

Indeed, $f'(x)=\frac{8(2-x)x^3}{3(x+1)(2x^2-3x+3)}$, which gives $x_{min}=0$

and since $f(3)=0$, we are done!

0

Note that you are trying to prove that $e^{q(x)} \geq \frac{p(x)}{p(x)-q(x)}$ for all $x$ smaller than the largest $0$ of $q(x)$...

  • 0
    You are right of course, but is this helpful? Is this true generally for any $p(x),q(x)$? – 2017-01-19
  • 0
    Divide both the numerator and the denominator of the right-hand term by $p(x)$, and write the result as an infinite series... – 2017-01-19
  • 0
    take the logarithm of both sides of the given inequality – 2017-01-19
  • 0
    Writing the infinite series gives me $\sum_{k=0}^{\infty}(q(x)/p(x))^k$. How do I compare this with $e^{q(x)} = \sum_{k=0}^{\infty}q(x)^k/k!$? – 2017-01-19
  • 0
    What are the signs involved, and the value of p(x)? – 2017-01-19
  • 0
    $q(x),p(x) \geq 0$ for $x \geq 0$ so $\frac{q(x)^k}{k!} \leq \frac{q(x)^k}{p(x)^k}$ for a fixed value of $x$ and $k$ large enough. – 2017-01-19
  • 0
    Dr Graubner, what should I do after taking the logarithm? – 2017-01-19
  • 0
    @Bill See my proof. – 2017-01-19