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I'M THE NEEDY ONE!!

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, 4 months ago

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$$3a^4 - 4a^3b + b^4$$

$$= a^4 + b^4 + 2a^4 - 4a^3b$$

$$= a^4 + b^4 - 2a^2b^2 + 2a^4 - 4a^3b + 2a^2b^2$$

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

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

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

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 $$\geq 0$$.

- 3 years, 4 months ago

THANKS A LOT!!

- 3 years, 4 months ago

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

- 3 years, 4 months ago