# Question about existence of solution for an inequality

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}}

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}}

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.

## 1 Answer

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.