Minimum value of x^2/(x-9)

I was asked to use the AM-GM inequality to solve this, I was thinking that I should express it into 2 fractions, and so i did, i tried to express x^2 as (x-9)^2 +18x -81 /x-9, though i couldn't find an answer.



Set $x-9=h$ $$\dfrac{x^2}{x-9}=\dfrac{(h+9)^2}h=h+\dfrac{9^2}h+18$$

If $h>0\iff x-9>0$ $$\dfrac{h+\dfrac{9^2}h}2\ge\sqrt{h\cdot\dfrac{9^2}h}=?$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.