New user? Sign up

Existing user? Sign in

Please tell me how i can prove that for any real numbers "a" and "b":

3({a}^4) - 4({a}^3)b + {b}^4 is equal to or greater than 0!!

Note by Jaiveer Shekhawat 3 years, 1 month ago

Sort by:

\( 3a^4 - 4a^3b + b^4 \)

\( = a^4 + b^4 + 2a^4 - 4a^3b \)

\( = a^4 + b^4 - 2a^2b^2 + 2a^4 - 4a^3b + 2a^2b^2 \)

\( = (a^2-b^2)^2 + 2a^2(a^2 + b^2 - 2ab) \)

\( = (a-b)^2(a+b)^2 + 2a^2(a-b)^2 \)

\( = (a-b)^2( 2a^2 + (a+b)^2) \)

Since first term is \( \geq 0 \) as its a square and the second term is \( \geq 0 \) as it a sum of squares. Therefore their product is also \( \geq 0 \).

Log in to reply

THANKS A LOT!!

As a and b are real no then if a = b then is eqal to 0. For a not eqal to b then the given value is greater than 0

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewest\( 3a^4 - 4a^3b + b^4 \)

\( = a^4 + b^4 + 2a^4 - 4a^3b \)

\( = a^4 + b^4 - 2a^2b^2 + 2a^4 - 4a^3b + 2a^2b^2 \)

\( = (a^2-b^2)^2 + 2a^2(a^2 + b^2 - 2ab) \)

\( = (a-b)^2(a+b)^2 + 2a^2(a-b)^2 \)

\( = (a-b)^2( 2a^2 + (a+b)^2) \)

Since first term is \( \geq 0 \) as its a square and the second term is \( \geq 0 \) as it a sum of squares. Therefore their product is also \( \geq 0 \).

Log in to reply

THANKS A LOT!!

Log in to reply

As a and b are real no then if a = b then is eqal to 0. For a not eqal to b then the given value is greater than 0

Log in to reply