I resulted a contradiction solving an inequality

Question

The problem is

"Find the range of $k$ when $x<1$ for the following inequality"

$$9k-6 \leq (3k-1)3^x$$

.

Isn't this equation always true because

$9k-6 \leq (3k-1)3^x < (3k-1)3 = 9k-3$

$9k-6<9k-3$

However, the answer sheet says it is $k\leq\frac{2}{3}$

Is there something wrong with my solution?

Answer 1

Your second inequality is wrong because $3k-1$ can be negative.

For $\frac{1}{3}\leq k\leq\frac{2}{3}$ it's obvious.

Let $k<\frac{1}{3}.$

Thus, we need $$3^x\leq\frac{3(3k-2)}{3k-1},$$ which is true because $$3^x<3<\frac{3(3k-2)}{3k-1}.$$ For $k>\frac{2}{3}$ we need $$3^x\geq\frac{3(3k-2)}{3k-1},$$ which is wrong for $$x<\log_3\frac{3(3k-2)}{3k-1}.$$ Id est, $\left(-\infty,\frac{2}{3}\right]$ is an answer.

Answer 2

You started by assuming that the inequality is true (for some $x<1,k$), and arrived at a conclusion which is always true (that is, for all $x,k$). This doesn't prove that the said inequality is true for all $k,x<1$.