Find the smallest value of $a+b^3$, where $a$ and $b$ are positive real numbers satisfying $ab=1$

Question

I am unsure as to where to go with this problem. Appreciate anyone who tries to help.

Answer 1

Hint: The constraint means that $b=a^{-1}$, so you need to minimize $f(a)=a+a^{-3}$ on an appropriate domain.

Note: You are probably looking for a local minimum since $f(a)$ increases without bound as $a\to 0^+$ or $a\to\infty$.

Answer 2

$a=\frac{1}{b}$

Then we have to minimize: $\frac{1}{b}+b^3$

The derivative occurs at $\frac{3b^4-1}{b^2}$ The zeros of the derivative are located at $\frac{1}{3^{\frac{1}{4}}}$and the negative of that number (but we are dealing with positive numbers. So the minimum occurs if $b= \frac{1}{3^{\frac{1}{4}}}$

Answer 3

$$1/b+b^3=3\cdot1/3b+b^3$$

Use https://en.m.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means

$$\dfrac{k\cdot1/kb+b^3}{k+1}\ge\sqrt[k+1]{(1/k)^kb^{3-k}}=?$$

Set $k-3=0$ to make the right hand side a constant

Answer 4

Since a , b are +ve hence use $W.A.M\ge W.GM $ on a and $3b^3$ with weightages as 1, and $\frac{1}{3}$. $$\frac{1.a+\frac{1}{3}(3b^3)}{1+\frac{1}{3}}\ge (a.(3b^3)^{\frac{1}{3}})^{\frac{1}{(1+\frac{1}{3})}}$$ $$\frac{a+b^3}{\frac{4}{3}}\ge (3^\frac{1}{3} a.b)^{\frac{3}{4}}$$ $$\therefore (a+b^3)\ge (\frac{4}{3})(3)^{\frac{1}{4}}$$ Minimum of $(a+b^3)$ is $4.3^{-{\frac{3}{4}}}$

Answer 5

The Lagrange multiplier method can be used to find the local maxima and minima of a function subject to equality constraints:

$$g(a,b,\lambda)=a+b^3-\lambda(ab-1)$$

$\dfrac {\partial g }{\partial a}=1-\lambda b=0\tag1$

$\dfrac {\partial g }{\partial b}=3b^2-\lambda a =0\tag2$

$\dfrac {\partial g}{\partial \lambda}=ab-1=0\tag3$.

From (1), $\lambda=\dfrac1b$. From (3), $a=\dfrac 1b$. Then from (2), $3b^2-\dfrac1{b^2}=0$, so $3b^4=1$.

There is only one real positive $b$ solving that last equation; can you take it from there?