Question

Inequality!

Answer 1

\[x-\frac{ x^2 }{ 2 }<\ln(1+x) ,\forall x>0\]

Answer 2

@myininaya @Abhisar @iambatman @Compassionate @eliassaab @mathmate @inkyvoyd @Joel_the_boss

Answer 3

Hint:
I suppose you have done McLaurin series and would be able to recognize that
ln(1+x) expands to \(x-\dfrac{x^2}{2!}+\dfrac{x^3}{3!}-\dfrac{x^4}{4!}+\dfrac{x^5}{5!}-...\), from which the first two terms are identical to the left-hand side.
So if you can prove that \(\dfrac{x^3}{3!}-\dfrac{x^4}{4!}+\dfrac{x^5}{5!}-... >0 ~~\forall x>0\) then you're done.

Answer 4

I thought about that expansion but works for |x|<1

Answer 5

Very good point, but don't dispose of it just yet.
If you visualize the graphs:
between 0 and 1,
f(x)=x-x^2/2! is increasing with a maximum at x=1.
but \(\forall\)x>1, f(x) is a decreasing function, while log(1+x) is increasing \(\forall x\in R\).
So your reasoning can be split into two two intervals, (0,1], and [1,\(\infty\)).

Answer 6

@mathmate what's next ?

Answer 7

My approach would be to complete the two separate parts of the proof, (0,1], and (1,\(\infty\)). First part using log, and second part using facts that one function is increasing, and the other decreasing.