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Given positive real variables $a$ and $b$, find the minimum of $$f(a,b)=\sqrt{\left(1+{1\over a}\right)\left(1+{1\over b}\right)}$$ subject to $a+b=\lambda$ where $\lambda$ is a constant . [ISI Sample Papers]

  1. Method $1$ : Substitute $b=\lambda - a$ and then compute ${\partial \over \partial a }f(a,b)$. But, the calculations get a bit messy.
  2. Method $2$ : Actually this is what I want to know. Is there an easier approach using some inequalities like the AM-GM inequality ? I tried this but was not able to got lost in between.
  3. Method $3$ : Lagrange multipliers. I have not tried this and kept it as a last option.

What is the best way to solve this problem ?

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You can write the term in the square root as $$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)=\frac{(a+1)(b+1)}{ab}=\frac{ab+a+b+1}{ab}=1+\frac{λ+1}{ab}$$ Now this term is minimized when $ab$ is maximized. So, your problem is equivalent to maximizing $ab$ subject to $a+b=λ$ i.e. to maximize $$a\cdot(λ-a)$$ Since it is a quadratic equation in $a$ with negative factor in front of $a^2$ the maximum is exactly the mean of the interval between the roots i.e. $a^*=\frac{λ}{2}$

Alternatively you could directly observe (or use) that the product $ab$ of two numbers with constant sum is maximized when the two numbers are equal.

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Noting that $ab\leq\frac{1}{4}(a+b)^2=\lambda^2/4$, we have $$ f(a,b)=\sqrt{\frac{1+a+b+ab}{ab}}=\sqrt{\frac{1+\lambda}{ab}+1}\geq\sqrt{\frac{4(1+\lambda)}{\lambda^2}+1} $$ with equality iff $a=b=\lambda/2$.

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Method 3: Lagrange multipliers.

Let $d>0$ and $U=\{(a,b);\;a>0,\;b>0\}$. Define $g:U\to\mathbb{R}$ and $\varphi:U\to \mathbb{R}$ by $g(a,b)=\left(\frac{1}{a+d}+1\right)\left(\frac{1}{b+d}+1\right)$ and $\varphi(a,b)=a+b$. Then, $$\nabla g(a,b)=-\left(\tfrac{1}{A^2}\left(\tfrac{1}{B}+1\right),\tfrac{1}{B^2}\left(\tfrac{1}{A}+1\right)\right)$$ where $A=a+d$ and $B=b+d$, and $\nabla\varphi(a,b)=(1,1)$.

It follows from the Lagrange Multipliers Method that:

$$\begin{align}(a,b)\in \varphi^{-1}(\lambda)\text{ is a critical point of } g\quad&\Longleftrightarrow\quad\nabla g(a,b)=\gamma \nabla \varphi(a,b) \text{ for some } \gamma\in\mathbb{R}\\\\ &\Longleftrightarrow \qquad\frac{1}{A^2}\left(\frac{1}{B}+1\right) =\frac{1}{B^2}\left(\frac{1}{A}+1\right)\\\\ &\Longleftrightarrow\qquad B+B^2=A+A^2\\\\ &\Longleftrightarrow\qquad B-A=A^2-B^2=-(A+B)(B-A)\\\\ &\Longleftrightarrow\qquad A=B\\\\ &\Longleftrightarrow\qquad a=b=\frac{\lambda}{2}\end{align}$$

So, $q=(\frac{\lambda}{2},\frac{\lambda}{2})$ is the unique critical point of $g$ in $\varphi^{-1}(\lambda)$.

It is possible to prove that $g$ has indeed a minimum at $q$ (see the argument here).

Therefore, for all $(a,b)\in\varphi^{-1}(\lambda)$ we have $$\left(\frac{1}{a+d}+1\right)\left(\frac{1}{b+d}+1\right)=g(a,b)\geq g\left(\tfrac{\lambda}{2},\tfrac{\lambda}{2}\right)=\left(\frac{1}{\tfrac{\lambda}{2}+d}+1\right)\left(\frac{1}{\tfrac{\lambda}{2}+d}+1\right)$$

Letting $d\to 0$ we conclude that $$f(a,b)^2=\left(\frac{1}{a}+1\right)\left(\frac{1}{b}+1\right)\geq \left(\frac{1}{\tfrac{\lambda}{2}}+1\right)\left(\frac{1}{\tfrac{\lambda}{2}}+1\right)=f\left(\tfrac{\lambda}{2},\tfrac{\lambda}{2}\right)^2$$ and thus the the minimum is attained at $(\tfrac{\lambda}{2},\tfrac{\lambda}{2})$.

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