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anemone
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Let a and b be positive real numbers such that ##a^5+b^3 \le a^2+b^2##.
Prove that ##a(a+b) \le 2##.
Prove that ##a(a+b) \le 2##.
Well hold on now. If you're going to find the maximum subject to the constraints, then you have to be concerned about the possibility that the maximum lies on the constraint, and the gradient is not zero there.bob012345 said:This was an interesting problem I enjoyed working on. Here is my solution.We are given $$a^5+b^3 \le a^2+b^2$$ and wish to show if that then $$a(a+b) \le 2$$
add ##a^2 + ab## to both sides
$$a^2 + ab + a^5+b^3 \le a^2 + ab + a^2+b^2$$ then move some terms over
$$a(a + b) \le 2a^2 + ab +b^2 - a^5 - b^3$$ the expression on the RHS is some function
$$f(a,b) = 2a^2 + ab +b^2 - a^5 - b^3$$ which we can find the maximum of. Taking partials we get
$$\frac{\partial f(a,b)}{\partial a} = 4a + b -5a^4 = 0$$ and
$$\frac{\partial f(a,b)}{\partial b} = a +2b -3b^2 = 0$$we get
$$ a = 3b^2 -2b$$ and $$b = 5a^4 -4a$$ substituting we get this beast
$$75a^7 -120a^4 -10a^3 + 48a +7 = 0$$ which has real positive solutions ##a=1## and ##a≈0.90173## the latter gives ##b≈-0.30812## which is not allowed so we have ##a=1## and therefore ##b=1##
this gives $$f(1,1) = 2a^2 + ab +b^2 - a^5 - b^3 = 2 + 1 + 1 - 1 -1 = 2$$ thus
$$a(a+b) \le 2$$
Mathcad 3D
View attachment 314293https://www.math3d.org/TRHrXnwZS
[\SPOILER]
I'm not finding the maximum of the LHS subject to the original constraint but the maximum of the entire function on the RHS and that is less than or equal to 2. I started with the constraint and transformed it into the desired condition on the LHS then showed that the resulting function on the RHS has a maximum of 2 given the original constraint. Look at the plot also please.Office_Shredder said:Well hold on now. If you're going to find the maximum subject to the constraints, then you have to be concerned about the possibility that the maximum lies on the constraint, and the gradient is not zero there.
To prove the inequality between two positive real numbers a and b, we need to show that a(a+b) ≤ 2. This can be done by using basic algebraic manipulations and properties of inequalities.
The inequality a(a+b) ≤ 2 is significant because it shows that the product of two positive real numbers a and b is always less than or equal to 2. This can have implications in various mathematical and scientific calculations and proofs.
Yes, for example, if we let a = 1 and b = 2, then a(a+b) = 1(1+2) = 3, which is greater than 2. This disproves the inequality. However, if we let a = 1 and b = 1, then a(a+b) = 1(1+1) = 2, which is equal to 2 and satisfies the inequality. This is just one example, but the inequality can be proven using various other values for a and b.
Yes, the inequality a(a+b) ≤ 2 is always true for positive real numbers a and b. This can be proven mathematically using the properties of inequalities and by considering various cases and values for a and b.
The inequality a(a+b) ≤ 2 can be applied in various real-world scenarios, such as in economics, physics, and engineering. For example, it can be used to determine the maximum amount of resources that can be allocated for a project, or the maximum load that a structure can withstand. It can also be applied in financial calculations, such as determining the maximum profit that can be made from a given investment.