Question

p,q,r,s are real numbers.show p^4 + q^4 >= 2(p^2)(q^2). Using that show p^4 + q^4 + r^4 + s^4 >= 4pqrs

Answer 1

use AM-GM inequality

Answer 2

first expression can be shown but second one cannot

Answer 3

\[\large \dfrac{p^4+q^4}{2} \ge \sqrt{p^4q^4}\]

Answer 4

\[\large \dfrac{p^4+q^4}{2} \ge p^2q^2\]

Answer 5

similarly

\[\large \dfrac{r^4+s^4}{2} \ge r^2s^2\]

Answer 6

right i did that.then what? I added them but how to get 4pqrs?

Answer 7

Adding them gives you :

\[\large \dfrac{p^4 + q^4}{2} + \dfrac{r^4+s^4}{2} \ge p^2q^2 + r^2s^2\]

Answer 8

\[\large p^4+q^4+r^4+s^4 \ge 4\left( \dfrac{p^2q^2 + r^2s^2}{2} \right)\]

\[\large p^4+q^4+r^4+s^4 \ge 4\left( \sqrt{p^2q^2*r^2s^2} \right)\]

Answer 9

oh right beautiful thanks again ganeshie8

Answer 10

nice @ganeshie8