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Thanx for pointing out the error. It is a decreasing function for x>e and not x>1 as i posted before

Here the log is taken to the base e

So, if x>e, then $\log_e {x} > \log_e{e} = 1$

Hence $1- \log_e{x} < 0$ for x>e

You must note that the proofs you suppy should be conclusive. There should be nothing left hanging in the air to be guessed or surmised.

Comparing $99^{100}$ and $100^{99}$ is equivalent to comparing $\sqrt [99] {99}$ and $\sqrt [100] {100}$

Now consider the function $f(x) = \sqrt [x] {x}$

$f'(x) = \frac{1 - \log x}{x^2}$

Hence for x>1, f is a decreasing function.

Thus, we deduce that $\sqrt [99] {99} > \sqrt [100] {100}$ .

To elaborate further, what is being used here is called the Sophie-Germain Identity

This factorization is often used to prove that a given expression is not a prime (unless of course one of the factors is 1)

$S = 1+(1+x)+(1+x^2)+...+(1+x+x^2+...+x^{n-1})$

$= x(1+x+x^2+...+x^{n-1}) - n$

$= x \ \left(\frac{x^n-1}{x-1} \right ) - n$

Hence

$S = x \ \left ( \frac{x^n-1}{(x-1)^2} \right) - \frac{n}{x-1}$

$\frac{\cos x}{\sin^2 x (\cos x - \sin x)}$

= $\frac{1}{\sin^2 x (1-\tan x)}$

= $\frac{1+\tan^2 x}{\tan^2 x (1-\tan x)}$

Hence the inequality becomes

$\frac{1+\tan^2 x}{\tan^2 x (1-\tan x)} \ge 8$

or $8\tan^3 x + 1 \ge 7 \tan^2x$ (since 1-tanx > 0)

$LHS = 8\tan^3 x+1 = 4\tan^3x + 4\tan^3 x + 1 \ge 3 \sqrt [3] {16\tan^6 x}$

= $6 \sqrt [3] {2} \tan^2 x$

It only remains to prove that $6 \sqrt [3] {2} \tan^2 x \ge 7 \tan^2 x$

or $6 \sqrt [3] {2} \ge 7$ which is found to be true by cubing both sides.

So, the inequality is a strict inequality

Pls see http://www.goiit.com/posts/list/trignometry-if-a-b-are-positive-quantities-and-if-41432.htm#209888

By any chance are you asking for ?

Are there some issues with the editor or is it just me.

When I try to edit the post, several extra spacings come in which makes the post unreadable.

Also, what is the criterion for locking the edit? I came back to the post half-an-hour later and I am not able to remove the spacings.

I may be making some mistake in formatting, so I would appreciate it if someone pointed it to me.

Thanx

Not quite sure what you need because it can be shown that

$\frac{a^a+b^b+c^c}{3} \ge \sqrt [3] {a^a b^b c^c} \ge (abc)^{\frac{a+b+c}{9}}$

If you could supply the choices, maybe we could manoeuver further!!

Notice that if

$\alpha$

is a root so is

$\frac{1}{\alpha}$

(Admin Sir, could you kindly see that the latex output comes inline with the text)

Both 1 and -1 are roots of the equation (That the product of roots is -1 also tells us that -1 is necessarily is a root)

Now $\Rightarrow 78(x^2-1)(x^4+x^2+1) = 133x(x^2-1) (x^2+1)$

$\Rightarrow 78(x^4+x^2+1) = 133x(x^2+1)$

(Since we have already noted that $x= \pm 1$ are roots)

$78x^2+ \frac{78} {x^2+1} = 133x$

$\Rightarrow 78(x^2+1)+\frac{78}{x^2+1} = 133x+78$

$\Rightarrow 133x+78 \ge 156 \ \text{(using AM-GM on} \ 78( x^2+1 + \frac{1}{x^2+1})$

$\Rightarrow x \ge \frac{78}{133}$

and since the reciprocal is also a root we must also have

$x\le \frac{133}{78}$

This means if x is a real root, then

$x, \frac{1}{x} \in \left( \frac{78}{133}, \frac{133}{78} \right )$

Now from rational roots theorem if x is a rational root of the equation, then

$x = \frac{p}{q}$

where p and q are both integer divisors of 78.

Here the real roots are positive. The factors of 78 to be considered are 1,2, 3, 6, 13, 26,39, 78

You can easily see that none of the fractions formed by these numbersother than $\frac{2}{3}$ satisfy the condition mentioned above. A quick check reveals that $\frac{2}{3}$ and hence $\frac{3}{2}$ are roots

Hence the rational roots are $\pm1$ , $\frac{2}{3}$ and $\frac{3}{2}$

Hence the sum of rational roots is $\boxed {\frac{13}{6}}$

Find $x, y \in \mathbb{R}$ such that

$x^2+20x+351 = \frac{2008}{y^2+30y+233}$

he asked for the sum of rational roots only

My attempt:

a+11b = 13

a+13b = 11

Eliminating b we get 2a = 169 - 121

Also 2b = 11 - 13

Hence 2(a+b) = 156 - 110

$\ge$ 1 and $\ge$ 2, and 11 - 13 $\ge$ 0 and and are of the same parity and hence the least difference is 2

give up!

I am unable to edit my earlier post, but I just want to say that when we say that n $\equiv$ r mod m+1 where r is prime to m+1 is the same as stating that n is prime to m+1.

So the condition is that n and m+1 are relatively prime