# A conjecture from Cauchy–Schwarz inequality

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?

• 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 $z_1=1,\ z_2=1, z_3=1$ Mar 12, 2013 at 9:59
• I think maybe if you cube the right-hand side instead of squaring, it might be correct. As it stands now it isn't. Mar 12, 2013 at 10:13
• @Arthur yes, i made a mistake Mar 12, 2013 at 10:38

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.

• yes, you are right. but is it true? Mar 12, 2013 at 10:43
• The proof is outlined above for non negative variables. Mar 12, 2013 at 11:08

Search for Lohwater's "Introduction to inequalities" (unpublished manuscript, was released by his wife).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.

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 $\alpha_1 = \alpha_2 = \alpha_3 = 1/3$.]

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

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 $(0.1+0.1+0.1)^{2}=0.09$

• i made a mistake, it should be on cube RHS Mar 12, 2013 at 10:39