Thread: Inequality

Originally Posted by pankaj

Prove that for

Define: f(x)=ln(1+x) and g(x)=1/{1+x^2} and use Mean value theorem - Wikipedia, the free encyclopedia on where .

Not sure... trying it...

Edit:

It's works!

What do you do for f(-1).

1st let f(x) = (1+x) ln(1+x) - Tan ' x

and then using calculus show that f(x) is increasing function in interval ]-1,0]

Appears to be easier said than done. I have been trying this approach but am not able to reach a definite conclusion

Originally Posted by pankaj

Appears to be easier said than done. I have been trying this approach but am not able to reach a definite conclusion

Read my post again(read the Cauchy mean value theorem).






Prove the last inequality.

the said inequality does hold in the given interval eg if x = -0.5 ln( 0.5) < Tan ' (0.5) / 0.5

Originally Posted by pankaj

Prove that for

Here is one way to do it. Multiply by 1+x (which doesn't change the inequality since 1+x>0). Then we want to show that



Since f(0) = 0 it will be sufficient to show that f(x) is decreasing in that interval. So we want to show that its derivative is negative. But



We want to show that for . To do that, repeat the previous argument. Since f'(0)=0 it is sufficient to show that f'(x) is increasing in that interval. So we want to show that its derivative is positive. But



which is clearly positive for x>–1.