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
6 years, 8 months ago

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Comments

3a44a3b+b4 3a^4 - 4a^3b + b^4

=a4+b4+2a44a3b = a^4 + b^4 + 2a^4 - 4a^3b

=a4+b42a2b2+2a44a3b+2a2b2 = a^4 + b^4 - 2a^2b^2 + 2a^4 - 4a^3b + 2a^2b^2

=(a2b2)2+2a2(a2+b22ab) = (a^2-b^2)^2 + 2a^2(a^2 + b^2 - 2ab)

=(ab)2(a+b)2+2a2(ab)2 = (a-b)^2(a+b)^2 + 2a^2(a-b)^2

=(ab)2(2a2+(a+b)2) = (a-b)^2( 2a^2 + (a+b)^2)

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

Siddhartha Srivastava - 6 years, 8 months ago

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THANKS A LOT!!

jaiveer shekhawat - 6 years, 8 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

Shyam Agarwal - 6 years, 8 months ago
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