Prove that for any real number 'a' we have 4a-a^4<=3.?

Question

<= means smaller or equal

Thank you.

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Edit:

If possible, I want a solution without differentiating if possible.

With elementary math workings.

Answer 1

well, we start a proof by manipulating the given and then formulate the proper solution. We can move all the terms to the other side which would give us a^4 - 4a + 3 >= 0.
We can apply factoring and synthetic division. By the remainder theorem, when a=1, a^4 - 4a + 3 would have a remainder of 0. So, it is a factor. By synthetic division, we get a^4 - 4a + 3 = (a-1)(a^3 + a^2 + a - 3). Again, we try to factor this expression again. by applying the same process, we arrive at a^3 + a^2 + a - 3 = (a-1)(a^2 + 2a + 3). plugging these factored expressions into the above, we arrive at:
a^4 - 4a + 3 = (a-1)^2(a^2 + 2a + 3). But we can't factor
(a^2 + 2a + 3) any more so we leave it that way. From the original a^4 - 4a + 3 >=0, we can say that
(a-1)^2(a^2 + 2a + 3) >=0. we know that a perfect square is always 0 and logically speaking a^2 > 2a except for a= -1,0. when a=0, it leaves 3 which is positive. when a= -1,
a^2 + 2a + 3 = 2 >0. Thus, positive x positive is always greater than or equal to 0. So, if you are to present your solution, the ideal way is working backwards. And start from the bottom.
The proof would go like this:
we know that this holds true: (a-1)^2(a^2 + 2a + 3) >=0 for any real value of a.
Expanding this, we arrive at:
a^4 - 4a + 3 >= 0.
Isolating the 3 in the other side gives us:
4a - a^4<= 3.
Therefore, it is true.
this style is commonly used in international mathematics contest, based on my experience as a delegate from an
asian english speaking country :D

Answer 2

Hi,

Moving everything to the right side gives the inequality

a^4 - 4a + 3 ≥ 0

Since a^4 is positive, this graph opens up and approaches ∞ as "a" --> ±∞. The first derivative, 4a³ - 4, solves to a = 1 when set equal to zero, indicating that its smallest value occurs at "a" = 1. When "a" = 1, a^4 - 4a + 3 = 0 and every other point would be higher than that minimum.

Therefore, a^4 - 4a + 3 ≥ 0 and 4a - a^4 ≤ 3 are both true.

I hope this helps!! :-)

Answer 3

Looking for the maximum of y = 4a - a^4.
4 - 4a^3 = 0
a^3 = 1
a = 1
4*1 - 1^4 = 3

Answer 4

If u diffrenciate 4a-a^4-3 and find the Extermoms u will find that the maximum is 3

Answer 5

WhAT?