Why is $\sqrt{ab} = \inf_{\lambda>0} \frac{a}{2\lambda}+\frac{\lambda b}{2}$?

Question

I saw this expression in a paper:

$\sqrt{ab} = \inf_{\lambda>0} \frac{a}{2\lambda}+\frac{\lambda b}{2}$

Can someone point me out why?

Answer 1

We will use the arithmetric-geometric inequality $\frac{x_1+x_2}{2} \geq\sqrt{x_1\cdot x_2}$:

$$ \inf_{\lambda > 0}\frac{a}{2\lambda}+ \frac{\lambda b}{2} = \inf_{\lambda > 0}\frac{\frac{a}{\lambda} + b\lambda}{2} \geq \inf_{\lambda > 0} \sqrt{\frac{a}{\lambda} \cdot b\lambda} = \inf_{\lambda > 0} \sqrt{ab} = \sqrt{ab} $$

So we know that this infimum is at least as big as $\sqrt{ab}$, but is it equal? We know that equality can be reached for a specific $\lambda$ s.t. $\frac{a}{\lambda}=b\lambda$, thus, we can assume that the infimum = minimum is exactly $\sqrt{ab}$. This infimum = minimum is reached for $\lambda = \sqrt{\frac{a}{b}} > 0$.

Answer 2

Let $$f(\lambda)=\frac{a}{2\lambda}+\frac{\lambda b}{2}$$ Then $$\frac{df}{d\lambda}=-\frac{a}{2\lambda^2}+\frac{b}{2}$$ and this equals zero when $$ \lambda=\lambda_{min}=\sqrt{\frac{a}{b}}$$ Finally $$f(\lambda_{min})=\sqrt{ab}$$

Answer 3

Assuming $a,b>0$

Let $f : \lambda \in \mathbb{R}_+^*\mapsto \dfrac{a}{2\lambda}+\dfrac{\lambda b}{2}$.

For all $\lambda>0$ we have $f'(\lambda)= -\dfrac{a}{2\lambda^2}+\dfrac{b}{2}$, hence:

$f'(\lambda) = 0 \iff \dfrac{b}{2} =\dfrac{a}{2\lambda^2} \iff \lambda = \sqrt{\dfrac{a}{b}}$ and you can check it is a global minimum:

$f'(\lambda)>0 \iff \lambda>\sqrt{\dfrac{a}{b}} $ so $f$ is decreasing on $\left(0,\sqrt{\dfrac{a}{b}}\right]$ and increasing on $\left[\sqrt{\dfrac{a}{b}},+\infty\right)$

Moreover $f\left(\sqrt{\dfrac{a}{b}}\right) = \dfrac{a}{2}\sqrt{\dfrac{b}{a}} +\sqrt{\dfrac{a}{b}} \dfrac{b}{2 } = 2\times\left( \dfrac{\sqrt{ab}}{2}\right) =\sqrt{ab} $

Therefore, $$\forall \lambda>0, f(\lambda)\geq\sqrt{ab}$$

Hence, $$\inf_{\lambda>0} f(\lambda) \geq \sqrt{ab}$$

But $\sqrt{ab} \in \{f(\lambda) \;\mathrm{s.t.} \;\lambda>0\}$ thus,

$$\inf_{\lambda>0} f(\lambda) \leq \sqrt{ab}$$

Finally:

$$\inf_{\lambda>0} f(\lambda) = \sqrt{ab}$$

Answer 4

The equation is clearly true if $a = 0$ and $b \geqslant 0,$ or if $b = 0$ and $a \geqslant 0.$

If $a > 0$ and $b > 0,$ then the equation is equivalent to $$ 2 = \inf_{\lambda > 0} \left( \frac{\sqrt{a/b}}{\lambda} + \frac{\lambda}{\sqrt{a/b}} \right), $$ and this in turn is equivalent to $$ 2 = \inf_{x > 0} \left( x + \frac1x\right), $$ which is easily proved, for instance by writing $$ x + \frac1x = 2 + \left( \sqrt{x} - \frac1{\sqrt{x}}\right)^2. $$

So it holds for all $a \geqslant 0$ and $b \geqslant 0.$