# How can I prove the inequality $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z}$?

For $x > 0$, $y > 0$, $z > 0$, prove:

$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z} .$$

I can see that this is true, I also checked it with a few numbers. But I guess that is not enough to prove it. So how can I prove it?

(You don't need to show the whole proof, I think just a hint will be enough. I simply don't know how to start.)

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Presumably, you want $>0$ rather than $\geq 0$, otherwise undefined –  Thomas Andrews Nov 3 '11 at 1:02
You are right, I fixed it. –  mcb Nov 3 '11 at 1:03
In view of the large number of very nice answers, perhaps the question should have been, is there any way to avoid proving the inequality? –  Gerry Myerson Nov 3 '11 at 5:23

Since $w+\frac{1}{w} \geq 2$ for all $w>0$, we have $$\begin{split} (x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) &= 3 + \frac{x}{y}+\frac{y}{x} + \frac{x}{z}+ \frac{z}{x} +\frac{y}{z}+\frac{z}{y} \\&\geq 3+2+2+2 = 9 \end{split}$$

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In general, the harmonic mean is at most the arithmetic mean, $$((1/n)(x_1^{-1}+\cdots+x_n^{-1}))^{-1}\le(1/n)(x_1+\cdots+x_n)$$ (assuming all quantities positive). Your question is the case $n=3$.

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This also follows from the Cauchy-Schwarz inequality:

Let $u=( x,y,z)$ and $v=( \frac{1}{x}, \frac{1}{y}, \frac{1}{z})$.

Then $u \cdot v \leq \|u\| \|v\|$ is exactly your inequality.

P.S. I personally find the other proofs mentioned above much simpler... And I would use the C-S inequality for numbers not vectors. But from my experience, many undergrad students know the C-S vector inequality and yet they don't know the HM-GM-AM inequality, this is the only reason I mentioned this proof :)

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Wow , thank you! I didn't know about this form of CS. –  A Googler Apr 20 at 12:38

$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z}$$ $\leftrightarrow (x+y+z)(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \geq 9$.

Using Cauchy Inequality, we have $x+y+z\geq 3\sqrt[3]{xyz}$ and $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\geq 3\sqrt[3]{\frac{1}{xyz}}$

$$\Rightarrow (x+y+z)\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \geq 3\sqrt[3]{xyz}\cdot 3\sqrt[3]{\frac{1}{xyz}} = 9$$.

Equality occurs when $x=y=z$.

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You can also brute-force it along with some careful re-grouping of terms:

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$

$\Leftrightarrow \frac{xy+xz+yz}{xyz}\geq \frac{9}{x+y+z}$

$\Leftrightarrow x^2 y + xyz + x^2 z + x y^2 + xyz + y^2 z + xyz + xz^2 + y z^2 \geq 9xyz$

$\Leftrightarrow x^2y-2xyz + yz^2 + x^2z-2xyz + yz^2 + xz^2-2xyz + xy^2 \geq 0$

$\Leftrightarrow y(x-z)^2 + z(x-y)^2 + x(y-z)^2 \geq 0$.

The last inequality holds (since $x,y,z>0$), therefore the first inequality holds.

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We don't need all above. We can use simple inequality

arithmetic mean ≥ geometric ≥ harmonic mean $$\frac{x+y+z}{3} ≥ \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \iff (x+y+z)\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) ≥ 9$$ hence proved..

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I tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post. For some basic information about writing math at this site see e.g. here, here, here and here. –  A.P. Apr 10 '13 at 7:19
–  A.P. Apr 10 '13 at 7:40
The right hand side of the $\iff$ should be a product, not a ratio. Also, in Heike's answer, the statement $w + 1/w \geq 2$ is the AM-GM inequality. –  Willie Wong Apr 10 '13 at 8:26