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we know the Cauchy–Schwarz inequality in $R^3$:

$$(x_1^2+x_2^2+x_3^2)(y_1^2+y_2^2+y_3^2)\geq(x_1y_1+x_2y_2+x_3y_3)^2$$

$$(x_1+x_2+x_3)(y_1+y_2+y_3)\geq(\sqrt{x_1y_1}+\sqrt{x_2y_2}+\sqrt{x_3y_3})^2$$

i guess the following inequlity exists too:

$$(x_1+x_2+x_3)(y_1+y_2+y_3)(z_1+z_2+z_3)\geq(\sqrt[3]{x_1y_1z_1}+\sqrt[3]{x_2y_2z_2}+\sqrt[3]{x_3y_3z_3})^3$$

is it true? how to prove?

Pedro
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Charles Bao
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  • you should add some infos like $x_i\geq 0$ what solutions would you accept? At first an heuristic one for the LHS you need $z_1+z_2+z_3=1$ so that the left hand side doesn't decrease, on the right hand side you need $z_1=1,\ z_2=1, z_3=1$ – Dominic Michaelis Mar 12 '13 at 09:59
  • I think maybe if you cube the right-hand side instead of squaring, it might be correct. As it stands now it isn't. – Arthur Mar 12 '13 at 10:13
  • @Arthur yes, i made a mistake – Charles Bao Mar 12 '13 at 10:38

5 Answers5

1

Lets look at a simple proof of Cauchy Schwarz inequality and see if we can extend in the direction you want.

Let $A^2 = \sum_{i=1}^n {a_i ^2}, \quad B^2 = \sum_{i=1}^n {b_i^2}$

Then $\sum_{i=1}^n \dfrac{a_i b_i}{AB} \le \sum_{i=1}^n \dfrac{1}{2}\left( \dfrac{a_i^2}{A^2} + \dfrac{b_i^2}{B^2} \right) = 1$
So $\sum_{i=1}^n {a_i b_i} \le AB$, or if you prefer $ A^2 B^2 \ge \left(\sum_{i=1}^n {a_i b_i} \right)^2$ which is Cauchy Schwarz.

Extending this, we have:

Let $A^3 = \sum_{i=1}^n {a_i ^3}, \quad B^3 = \sum_{i=1}^n {b_i^3}, \quad C^3 = \sum_{i=1}^n {c_i^3}$

Then $\sum_{i=1}^n \dfrac{a_i b_i c_i}{ABC} \le \sum_{i=1}^n \dfrac{1}{3} \left( \dfrac{a_i^3}{A^3} + \dfrac{b_i^3}{B^3} + \dfrac{c_i^3}{C^3} \right) = 1$

So $\sum_{i=1}^n {a_i b_i c_i} \le ABC$, or if you prefer, $A^3B^3C^3 \ge \left(\sum_{i=1}^n {a_i b_i c_i} \right)^3$ is the extended version. Essentially your RHS needs to be cubed, not squared.

Macavity
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Search for Lohwater's "Introduction to inequalities" (unpublished manuscript, was released by his wife). It is almost 200 pages full of all sorts of elementary techniques for proving inequalities.

vonbrand
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Found in Hewitt & Stromberg, Real and Abstract Analysis, (13.26) Exercise, Generalized Hölder's Inequality

Let $\alpha_1, \alpha_2, \dots \alpha_n$ be positive real numbers such that $\sum_{j=1}^n \alpha_j=1$. For nonnegative functions $f_1, f_2, \dots, f_n \in L_1(X,\mu)$, we have $$ f_1^{\alpha_1}f_2^{\alpha_2}\cdots f_n^{\alpha_n} \in L_1(X,\mu), $$ and $$ \int_X\big(f_1^{\alpha_1}f_2^{\alpha_2}\cdots f_n^{\alpha_n}\big) d\mu \le \|f_1\|_1^{\alpha_1}\|f_2\|_1^{\alpha_2}\cdots \|f_n\|_1^{\alpha_n} . $$

[Let $X$ have three points, each of measure $1$. Let $n=3$ and $\alpha_1 = \alpha_2 = \alpha_3 = 1/3$.]

GEdgar
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It's just another form of Holder's inequality, which generalises Cauchy-Schwarz.

Potla
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-1

It is false. $x_{1}=x_{2}=x_{3}=y_{1}=y_{2}=y_{3}=z_{1}=z_{2}=z_{3}=0.1$ Then LHS is $0.3*0.3*0.3=0.027$ RHS is $(0.1+0.1+0.1)^{2}=0.09$

user66380
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