Suggestions on an inequality?

Question

I want the following inequality hold: $$ (\xi_1a_1+\xi_2a_2+\xi_3a_3)(a_4+a_5+a_6) < (a_1+a_2+a_3)(\eta_1a_4+\eta_2a_5+\eta_3a_6) $$ where $a_1, ..., a_6>0$ and $0 \leq \xi_i < \eta_i$ for $i=1,2,3$.

Is this inequality true or false, and when this inequality can be guaranteed to hold? Any suggestions?

Answer 1

Not necessarily true, as in there are $a_1,\ldots, a_6$ and $\xi_1,\xi_2, \xi_3$ and $\eta_1,\eta_2,\eta_3$ that meet the conditions of your hypothesis and yet the (strict) inequality $(\xi_1a_1+\xi_2a_2+\xi_3a_3)(a_4+a_5+a_6) > (a_1+a_2+a_3)(\eta_1 a_4+\eta_2 a_5 +\eta_3 a_6)$ holds.

Suppose

1. $a_1=a_4=a_5=a_6 =1$ and $a_2=a_3 = \epsilon$

2. $\xi_1 = (1-\delta)$ for some small $\delta >0$; $\xi_2=\xi_3 = \epsilon$ for some small $\epsilon >0$

3. $\eta_1=1$, $\eta_2=\eta_3= 2\epsilon$.

Then the conditions of your hypothesis are met.

However, $(\xi_1a_1+\xi_2a_2+\xi_3a_3)(a_4+a_5+a_6) \ge 2.75$ if $\delta <.1$

Yet $(a_1+a_2+a_3)(\eta_1 a_4+\eta_2 a_5 +\eta_3 a_6) \le 1.25$ if $\epsilon < .1$.

Answer 2

Note that

$$(\xi_1a_1+\xi_2a_2+\xi_3a_3)(a_4+a_5+a_6) < (a_1+a_2+a_3)(\eta_1a_4+\eta_2a_5+\eta_3a_6)$$

$$\iff \frac{(\xi_1a_1+\xi_2a_2+\xi_3a_3)}{(a_1+a_2+a_3)} < \frac{(\eta_1a_4+\eta_2a_5+\eta_3a_6)}{(a_4+a_5+a_6)}\iff \bar \xi<\bar\eta$$

that is a Weighted arithmetic mean which is true in general for $a_1+a_2+a_3=a_4+a_5+a_6$.

To show that let assume

• $\eta_i=\xi_i+d_i \quad d_i>0$

therefore

$$\frac{(\eta_1a_4+\eta_2a_5+\eta_3a_6)}{(a_4+a_5+a_6)}=\frac{(\xi_1a_4+\xi_2a_5+\xi_3a_6)}{(a_4+a_5+a_6)}+\frac{(d_1a_4+d_2a_5+d_3a_6)}{(a_4+a_5+a_6)}=$$

$$=\bar \xi + \bar d>\bar \xi =\frac{(\xi_1a_1+\xi_2a_2+\xi_3a_3)}{(a_1+a_2+a_3)}$$

otherwise in general the inequality is not true and we can find counterexamples as shown also in Mike answer.