Minimum value of $\sqrt{(1+1/y)(1+1/z)}$

Question

If $y,z > 0$ and $y + z = c$ where $c$ is a constant, then what's the minimum value of

$$\sqrt{\left(1+\frac1y\right)\left(1+\frac1z\right)}$$

I am having a hard time solving this.

Answer 1

Since $f(x) = \sqrt{x}$ is bijective, it suffices to minimize what's inside the radical.

Now by the Cauchy Schwarz Inequality,

$\left(1+ \frac{1}y \right) \left( 1 + \frac{1}z \right) \ge \left(1+ \frac{1}{\sqrt{yz}} \right)^2$.

To minimize this, we need to maximize $yz$ which by AM GM, occurs when $y = z$. Therefore, the minimum is $\frac{c+2}{c}$.

Answer 2

Here's an elementary proof: $$\sqrt{\left(1+\frac1y\right)\left(1+\frac1z\right)}=\sqrt{1+\frac1y+\frac1z+\frac1{yz}}=\sqrt{1+\frac{y+z+1}{yz}}=\sqrt{1+\frac{1+c}{yz}}$$ So we have a minimum exactly when $yz$ is maximal, but $$0\le(y-z)^2\Longleftrightarrow4yz\le(y+z)^2=c^2$$ We see the maximum value is attained only when equality holds, i.e. $0=(y-z)^2\Longleftrightarrow y=z=\dfrac c2$, so the minimum is $$\sqrt{\left(1+\frac2c\right)\left(1+\frac2c\right)}=1+\frac2c$$

Answer 3

This can also be taken on using single-variable calculus by using the constraint to insert $ \ z \ = \ c \ - \ y \ $ into

$$ \left[ \ \frac{y \ (c-y) \ + \ (c-y) \ + \ y \ + \ 1}{y \ (c-y)} \ \right]^{1/2} \ = \ \left[ \ \frac{1 \ + \ c \ + \ cy \ - \ y^2}{y \ (c-y)} \ \right]^{1/2} \ \ . $$

Differentiating with respect to $ \ y \ $ , simplifying the resulting expression, and setting that equal to zero yields

$$ \frac{1}{2} \ \frac{\sqrt{y \ (c-y)} \ \cdot \ (1+c) \ (c - 2y)}{[1+c+cy-y^2]^{1/2} \ \cdot \ y^{3/2} \ (c-y)^{3/2}} \ = \ 0 \ \ . $$

Since the domain for the function is $ \ 0 \ < \ y \ < \ c \ $ , the only value that makes the numerator zero is $ \ y \ = \ \frac{c}{2} \ $ , from which it follows that $ \ z \ = \ \frac{c}{2} \ $ and that the minimal value for the function is $ \ \ \ \frac{c + 2}{c} \ $ , as user2345215 calculates. (It is not difficult to show that this is a local minimum by varying $ \ y \ $ slightly.)

[Just to demonstrate this, if we let $ \ y \ = \ \frac{c}{2} \ + \ \delta \ $ and $ \ z \ = \ \frac{c}{2} \ - \ \delta \ $ , the value of the function becomes

$$ \sqrt{\frac{ \ [ \ \frac{c}{2} \ + \ 1 \ ]^2 \ - \ \delta^2}{ \ ( \frac{c}{2} )^2 \ - \ \delta^2}} \ \ ; $$ since the first term in the numerator is larger than that of the denominator, subtracting $ \ \delta^2 \ $ from it reduces it proportionally less. Thus, the ratio under the square-root becomes larger.]

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That was a bit unpleasant -- Lagrange-multipliers are somewhat easier to work with. Since $ \ y \ $ and $ \ z \ $ are positive, as are the square-roots in this function, we can just extremize $ f(y,z) $ $ = \left(1 + \frac{1}{y} \right) \ \left(1 + \frac{1}{z} \right) \ $ , subject to the constraint $ \ g(y,z) \ = \ y \ + \ z \ - \ c \ $. For the Lagrange equations $ \ \nabla f \ = \ \lambda \ \nabla g \ $ , we have

$$ -\frac{1}{y^2} \ - \ \frac{1}{y^2z} \ = \ \lambda \ \cdot \ 1 \ \ , \ \ -\frac{1}{z^2} \ - \ \frac{1}{yz^2} \ = \ \lambda \ \cdot \ 1 \ \ $$

$$ \Rightarrow \ \ \frac{z \ + \ 1}{y^2z} \ = \ \frac{y \ + \ 1}{yz^2} \ \ \Rightarrow \ \ z^2 \ + \ z \ = \ y^2 \ + \ y \ \ . $$

Now inserting $ \ z \ = \ c \ - \ y \ $ , we obtain

$$ (c \ - \ y)^2 \ + \ (c \ - \ y ) \ = \ y^2 \ - \ y \ \ \Rightarrow \ \ c \ (c + 1) \ = \ 2y \ (c + 1) \ \ , $$

from which we again find $ \ y \ = \ z \ = \ \frac{c}{2} \ $ .

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We can also really just exploit the symmetry of the function itself: since it is unchanged by "exchanging" the variables $ \ y \ $ and $ \ z \ $ , the function has "diagonal symmetry" about the line $ \ \ y \ = \ z \ $ . This line intersects the constraint line $ \ y \ + \ z \ = \ c \ $ at $ \ y \ = \ z \ = \ \frac{c}{2} \ $ , so an extremum for the function lies there (found to be the minimum).