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This question already has an answer here:

Let $x,y,z$ be positive real numbers satisfying $x^2+y^2+z^2=1$. Prove that :

$$\frac {1}{x} + \frac {1}{y}+\frac {1}{z} \geq 3{\sqrt{3}}.$$

I derived the equality case easily. I was able to prove the inequality with the help of Lagrange Multipliers, which made it look very easy. Is there any other way to prove the same inequality without calculus ? I tried AM-GM and Cauchy-Schwarz but could not find a proper set of values to apply these on so as to obtain the inequality.

Any help would be appreciated . :)

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marked as duplicate by Martin R, Community Jan 18 '17 at 16:19

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  • $\begingroup$ Maybe a hint: $\frac{xy + yz + xz}{3} \geq \sqrt{3x^2y^2z^2}$. $\endgroup$ – Cehhiro Jan 18 '17 at 15:54
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Simplest is to use the inequality between the harmonic mean and quadratic mean:

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

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  • $\begingroup$ Yeah.. I think that is the simplest solution..... +1..... $\endgroup$ – user399078 Jan 18 '17 at 15:58
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First we have $\frac{xy + yz + xz}{3} \geq \sqrt{3x^2y^2z^2} = \sqrt{3(xy)(yz)(zx)}$.

Substitute $a = xy$, $b = yz$, $c = xz$.

$$ \frac{a+b+c}{3} \geq \sqrt[3]{abc} $$

Finally it's necessary to show $\sqrt[3]{abc} \geq \sqrt{3\ abc}$, but without calculus I lack the tools to prove it. If anyone wants to pick it up from here, it's open.

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  • $\begingroup$ Just trying to help... $\sqrt[3]{abc} \geq \sqrt{3\ abc}$ is equivalent to $abc\le\frac{1}{3^3}$, which can be deducted from $\frac{a^2+b^2+c^2}{3}\ge\sqrt[3]{a^2b^2c^2}$ $\endgroup$ – Momo Jan 18 '17 at 16:18
  • $\begingroup$ @Momo Thanks! I'm a potato with this area of Mathematics. I appreciate the help. (I'll leave the answer as is to not take undeserved credit.) $\endgroup$ – Cehhiro Jan 18 '17 at 16:24