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How can I prove $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq 27$, given that $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9$ and $x+y+z=1$.

I've already tried using that: $\frac{1}{x} +\frac{1}{y} +\frac{1}{z}\geq 9$ But I can't seem to manipulate that to prove the above.

Stefan4024
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Azza
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  • The title isn't supposed to be the first line of your question. – Git Gud Sep 28 '13 at 10:14
  • @GitGud :I don't get it, what is wrong? – jimjim Sep 28 '13 at 10:28
  • Can you explain your question a bit more? What have you tried already? – user93089 Sep 28 '13 at 10:30
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    @Arjang just as I said: the title isn't supposed to be the first line of your question. Effectively your 'question' is $$\text{Given that }(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9 \text{ and }x+y+z=1 $$ which isn't a question at all. – Git Gud Sep 28 '13 at 10:31

2 Answers2

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Cauchy-Schwarz inequality tells us that \begin{equation} \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2 \leq 3\times \left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\qquad (\star) \end{equation} But the left-hand term is $\geq 9^2$, so $$ \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \geq \frac{9^2}{3} = 27. $$

Note that in this particular case, it very is easy to prove $(\star)$. Let $$ A = 3 \left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right),\qquad B = \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2. $$

We need to show that $A \geq B$. Taking their difference, \begin{align} A - B & = \frac{1}{x^2}+\frac{1}{y^2} - 2\frac{1}{xy} + \frac{1}{y^2} + \frac{1}{z^2} - 2\frac{1}{yz} + \frac{1}{z^2}+\frac{1}{x^2} - 2\frac{1}{zx}\\ & = \left(\frac{1}{x}-\frac{1}{y}\right)^2 + \left(\frac{1}{y}-\frac{1}{z}\right)^2 + \left(\frac{1}{z}-\frac{1}{x}\right)^2\\ & \geq 0. \end{align}

Siméon
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You may also try this :

Apply AM-HM inequality on the set $\{x^2,y^2,z^2\}$ :

$$\frac {x^2+y^2+z^2}{3} \geq \frac {3}{\frac {1}{x^2}+\frac {1}{y^2}+\frac{1}{z^2}}$$

$$\implies \big(x^2+y^2+z^2\big)\big({\frac {1}{x^2}+\frac {1}{y^2}+\frac{1}{z^2}}\big) \geq 9$$

$$\implies {\frac {1}{x^2}+\frac {1}{y^2}+\frac{1}{z^2}} \geq \frac{9}{x^2+y^2+z^2}\,\,\,(♦)$$

Now all that remains is to prove that $$x^2+y^2+z^2 \leq \frac{1}{3} \,\,\,(♣)$$

with the constraint that $$x+y+z=1 $$

I am assuming that you are skilled enough to prove $(♣)$. (You can use Lagrange Multipliers or some other technique.)

Plug $(♣)$ into $(♦)$ to get :

$$ {\frac {1}{x^2}+\frac {1}{y^2}+\frac{1}{z^2}}\geq \frac{9}{1/3} = 27$$

$$HENCE \,\,\,\, PROVED$$ NOTE : Since $x^2+y^2+z^2$ is in the denominator (♦) , the sign of (♣) flips/reverts , thus proving the desired inequality.