Inequality problem $\frac 1x + \frac 1y + \frac 1z > 5$ prove

Question

Prove that:

$$\frac 1x + \frac 1y + \frac 1z > 5$$

where $x+y+z=1$, $x, y, z$ are real numbers not equal $0$ and $x\neq y \neq z $

Answer 1

Note: This somewhat longish answer contains several proofs of the inequality. I'm leaving things in the order in which they accreted, but I recommend skipping to the boldface "added yet later" section for the simplest of the proofs here. I also highly recommend Macavity's answer as the simplest of all.

Using AGM twice, we have

$${1\over x}+{1\over y}+{1\over z}\ge3\sqrt[3]{1\over xyz}={3\over\sqrt[3]{xyz}}\ge{3\over\left(x+y+z\over3\right)}={9\over x+y+z}=9\gt5$$

Added later: Here is a second, completely different proof, which gets the requested inequality (with a $5$) without getting the stronger inequality with a $9$.

The requirement that $x+y+z=1$ with $x,y,z\gt0$ means that $(\sqrt x,\sqrt y,\sqrt z)$ lies on the unit sphere.. This means we can parameterize the variables, using spherical coordinates, as

$$\begin{align} x&=\sin^2\theta\cos^2\phi\\ y&=\sin^2\theta\sin^2\phi\\ z&=\cos^2\theta \end{align}$$

Then

$$\begin{align} {1\over x}+{1\over y}+{1\over z} &={1\over\sin^2\theta\cos^2\phi}+{1\over\sin^2\theta\sin^2\phi}+{1\over\cos^2\theta}\\ &={\sin^2\phi+\cos^2\phi\over\sin^2\theta\sin^2\phi\cos^2\phi}+{1\over\cos^2\theta}\\ &={1\over\sin^2\theta\sin^2\phi\cos^2\phi}+{1\over\cos^2\theta}\\ &={4\over\sin^2\theta(2\sin\phi\cos\phi)^2}+{1\over\cos^2\theta}\\ &={4\over\sin^2\theta\sin^22\phi}+{1\over\cos^2\theta}\\ &\ge4+1\\ &=5 \end{align}$$

Added yet later: Here is a third, fairly simple proof.

The conditions on $x$, $y$, and $z$ imply $0\lt x,y,z\lt1$, so we can write $x=1-u$, $y=1-v$, and $z=1-w$ with $0\lt u,v,w\lt1$. The equation $x+y+z=1$ translates into $u+v+w=2$. Thus

$$\begin{align} {1\over x}+{1\over y}+{1\over z} &={1\over1-u}+{1\over1-v}+{1\over1-w}\\ &=(1+u+u^2+\cdots)+(1+v+v^2+\cdots)+(1+w+w^2+\cdots)\\ &\gt(1+u)+(1+v)+(1+w)\\ &=3+(u+v+w)\\ &=5 \end{align}$$

Remarks: If you include Macavity's extremely simple proof, we now have three proofs of the requested inequality that don't give anything stronger without additional work. It's worth noting that Macavity's approach can be strengthened by $1$ with just a small twist: If we order $0\lt x\le y\le z\lt1$, then $x+y+z=1$ implies $x\le{1\over3}$ and $y\le{1\over2}$, which gives

$${1\over x}+{1\over y}+{1\over z}\gt3+2+1=6$$

It might be entertaining to see if there are any approaches that naturally (whatever "naturally" means) give the inequality with a $7$ but not an $8$, or an $8$ but not the $9$. Update: Macavity has given proofs that do this in comments below.

Answer 2

You have $$ (x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq 9$$ To see this just develop everything and use $a/b+b/a \geq 2$. Another variant is to use Cauchy Schwarz.

Now, if $x+y+z = 1$ it follows that $1/x+1/y+1/z \geq 9$.

Answer 3

We can even prove that

$$ \frac 1x + \frac 1y + \frac 1z \ge 9\;. $$

Consider the intersection of the domain with the cube $\left[\frac19,\frac89\right]^3$. This is a compact set on whose boundary the left-hand side is clearly $\ge9$. The left-hand side attains its minimum on this compact set either on the boundary or at a stationary point in the interior. Using a Lagrange multiplier, we find as a condition for stationary points

$$ \frac1{x^2}=\frac1{y^2}=\frac1{z^2}=\lambda $$

and thus $x=y=z=\frac13$. Thus the minimum is $9$.

Answer 4

Given the weak bound, here is a simpler way: the smallest among the three variables must be less than $\frac13$ for obvious reasons, and the other two are less than $1$ as all variables are positive. Hence the reciprocals sum to more than $3+1+1=5$.

Answer 5

The inequality that joriki have given has a general form, in case you wanted to know : $$ \sum_{i=1}^n \frac {a_i^2}{b_i}\ge \frac{(\sum_{i=1}^n a_i)^2}{\sum_{i=1}^n b_i} $$ with $n$ being an integer and $\{a_1;a_2;...:a_n\}$ and $\{b_1;b_2;...:b_n\} $ being positve real numbers. The inequality can be prove by induction .