Question about existence of solution for an inequality

Question

Question

Let $a_1,a_2,b_1,b_2$ and $x_1,x_2, y_1,y_2$ be non-negative numbers. Assume that:

1. $(2 x_1+a_2)(2 y_1+b_2) > (2 a_1+a_2)(2 b_1+b_2)$

2. $(2 a_1+x_2)(2 b_1+y_2) > (2 a_1+a_2)(2 b_1+b_2)$

Does there exist non-negative numbers $z_1,z_2,w_1,w_2$ such that:

1. $z_1+z_2=x_1+x_2$,
2. $w_1+w_2=y_1+y_2$,
3. $(2 z_1+z_2)(2 w_1+w_2) > (2 a_1+a_2)(2 b_1+b_2)$

If exists, then the statement still holds true if we replace $2$ by any $\lambda >0$?

COMMENTS

• I feel that the answer is yes. For example, we can choose $z_1=x_1,z_2=x_2,w_1+y_1,w_2=y_2$ in the simple case when $x_1\geq a_1,x_2 \geq a_2,y_1 \geq b_1,y_2 \geq b_2$.

• I try to make a simple calculation by finding $z_1,z_2,w_1,w_2$ as follows:

$$\begin{align} 2z_1+z_2=(2x_1+a_2)^\theta(2a_1+x_2)^{1-\theta}\\ 2w_1+w_2=(2y_1+b_2)^\theta (2b_1+y_2)^{1-\theta} \end{align}$$ for some $\theta \in (0,1)$. Combining with equations 1 and 2, I'm able to compute $z_1,z_2,w_1,w_2$. But I cannot show that they are non-negative.

EDIT

Thanks a lot for the answer by S.Dolan. Now I change a bit on the question as follow

Let $a_1,a_2,b_1,b_2$ and $x_1,x_2, y_1,y_2$ be non-negative numbers. Assume that:

1. $e^{-x_1+\frac{a_2}{2}}+e^{-y_1+\frac{b_2}{2}} < e^{-a_1+\frac{a_2}{2}}+e^{-b_1+\frac{b_2}{2}}$

2. $e^{-x_2+\frac{a_1}{2}}+e^{-y_2+\frac{b_1}{2}} <e^{-a_2+\frac{a_1}{2}}+e^{-b_2+\frac{b_1}{2}}$

Does there exist non-negative numbers $z_1,z_2,w_1,w_2$ such that:

1. $z_1+z_2=x_1+x_2$,

2. $w_1+w_2=y_1+y_2$,

3. $e^{-z_1+\frac{z_2}{2}}+e^{-w_1+\frac{w_2}{2}} < e^{-a_1+\frac{a_2}{2}}+e^{-b_1+\frac{b_2}{2}}$

4. $e^{-z_2+\frac{z_1}{2}}+e^{-w_2+\frac{w_1}{2}} <e^{-a_2+\frac{a_1}{2}}+e^{-b_2+\frac{b_1}{2}}$

If exists, then the statement still holds true if we replace $\dfrac{1}{2}$ by any $1>\lambda >0$?

Btw any comments and discussions are very welcome.

Thanks in advance.

Answer 1

This is not generally true.

Let $y_1=y_2=0,a_i=b_i=1$ and let the $x_i$ be any sufficiently large numbers.