# Inequality with one equality constraint … [duplicate]

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 . :)

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

• Maybe a hint: $\frac{xy + yz + xz}{3} \geq \sqrt{3x^2y^2z^2}$. – Cehhiro Jan 18 '17 at 15:54

Simplest is to use the inequality between the harmonic mean and quadratic mean:

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

• Yeah.. I think that is the simplest solution..... +1..... – user399078 Jan 18 '17 at 15:58

First we have $\frac{xy + yz + xz}{3} \geq \sqrt{3x^2y^2z^2} = \sqrt{3(xy)(yz)(zx)}$.

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

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

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

• Just trying to help... $\sqrt{abc} \geq \sqrt{3\ abc}$ is equivalent to $abc\le\frac{1}{3^3}$, which can be deducted from $\frac{a^2+b^2+c^2}{3}\ge\sqrt{a^2b^2c^2}$ – Momo Jan 18 '17 at 16:18
• @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.) – Cehhiro Jan 18 '17 at 16:24