# For positive real numbers $a_1, a_2, b_1, b_2$ holds $a_1 + a_2 > b_1 + b_2$ and $a_1a_2 < b_1b_2$. Prove that $a_1 > b_1 > b_2 > a_2$.

For positive real numbers $$a_1, a_2, b_1, b_2$$ satisfying $$a_1 > a_2, b_1 > b_2$$ holds \begin{align} a_1+a_2 &> b_1+b_2 \tag{1}\label{eq1} \\ a_1a_2 &< b_1b_2 \tag{2}\label{eq2} \end{align}

Prove that then necessarily $$a_1 > b_1 > b_2 > a_2$$.

This is what I have tried

Introduction. Since $$a_1 > a_2, b_1 > b_2$$ then the largest number from $$a_1, a_2, b_1, b_2$$ can be $$a_1$$ or $$b_1$$. Also the smallest number can be $$a_2$$ or $$b_2$$.

1.) If $$a_1 = b_1$$ then it follows from $$(1)$$ that $$a_2 > b_2$$. However, then $$(2)$$ doesn't hold. $$a_1 \neq b_1$$.

2.) If $$a_1 < b_1$$. Let $$a_1 = x - \varepsilon,\, b_1 = x + \varepsilon,\, \varepsilon > 0$$. If we substitute this into $$(1)$$ we get $$a_2 > b_2 + 2\varepsilon \implies a_2 > b_2$$. From discussion in the introduction and from $$a_1 > a_2$$ it then follows $$b_1>a_1>a_2>b_2$$.

3.) If $$a_1 > b_1$$ then for $$(2)$$ to hold it must necessarily $$a_2 < b_2$$ and $$b_1 > a_2$$. So it follows that $$a_1>b_1>b_2>a_2$$. For example for $$a_1 = 5,\, a_2 = 1,\, b_1 = 3,\, b_2 = 2$$ both inequalities $$(1)$$ and $$(2)$$ hold.

As you can see in 2.) I haven't found some contradiction that would show that $$b_1>a_1>a_2>b_2$$ is not possible. Can someone help me with this?

Let $$f(x)=(x-a_1)(x-a_2)=x^2-(a_1+a_2)x+a_1a_2$$ and $$g(x)=(x-b_1)(x-b_2)=x^2-(b_1+b_2)x+b_1b_2.$$ By the conditions on $$a_i$$ and $$b_i$$, we see that $$f(x) for every $$x>0$$. Hence $$(b_1-a_1)(b_1-a_2)=f(b_1) and $$(b_2-a_1)(b_2-a_2)=f(b_2) Using $$a_1>a_2$$ and $$b_1>b_2$$, the two inequalities above show that $$a_1>b_1>b_2>a_2$$.

If we know that $$a_1 > a_2 > 0$$ and $$b_1 > b_2 > 0$$ you can do the following $$a_1 + a_2 > b_1 + b_2 \rightarrow (a_1 + a_2)^2 > (b_1 + b_2)^2 \rightarrow a_1^2 + a_2^2 + 2a_1a_2 > b_1^2 + b_2^2 + 2b_1b_2.$$ Observe that $$a_1a_2 < b_1 b_2 \rightarrow -4a_1a_2 > -4b_1b_2$$ and add everything $$a_1^2 + a_2^2 - 2a_1a_2 > b_1^2 + b_2^2 - 2b_1b_2 \rightarrow (a_1 - a_2)^2 > (b_1 - b_2)^2$$ Since $$a_1 > a_2$$ and $$b_1 > b_2$$ $$a_1 - a_2 > b_1 - b_2.$$ Now adding it to the first inequality we can get $$2 a_1 > 2b_1 \rightarrow a_1 > b_1.$$ By assumption you know that $$b_1 > b_2$$,thus $$a_1 > b_1 > b_2.$$ The last one is a bit tricky. Use that $$a_1 > b_1 \rightarrow \frac{1}{a_1} < \frac{1}{b_1}$$ and multiply the inequality $$a_1 a_2 < b_1 b_2$$ to get $$a_2 < b_2$$ that finishes the proof $$a_1 > b_1 > b_2 > a_2.$$

Well whenever I see inequalities involving $$m + n$$ and $$mn$$ I immediately think if squaring.

$$a_1 + a_2 > b_1 + b_2 \implies$$

$$(a_1+a_2)^2 > (b_1+b_2)^2 \implies$$

$$a_1^2 + 2a_1a_2 + a_2^2>b_1^2 + 2b_1b_2 + b_2^2$$

And we have $$a_1a_2 < b_1b_2$$ so $$-a_1a_2 > -b_1b_2$$ so

$$a_1^2 - 2a_1a_2 + a_2^2 >b_1^2 -2b_1b_2 + b_2^2$$ so

$$(a_1 - a_2)^2 > (b_1-b_2)^2$$ so

$$a_1 - a_2 > b_1 - b_2$$ and $$a_1 + a_2 > b_1 + b_2$$.

So add those and we get

$$(a_1 - a_2)+(a_1+a_2) > (b_1 - b_2)+(b_1+b_2)$$ so $$a_1 > b_1$$.

......

So we have $$a_1 > b_1 >b_2$$ but where does $$a_2$$ fit in?

Well we have $$a_1a_2 < b_1b_2$$ and to isolate the $$a_2$$ and $$b_2$$ we have $$\frac {a_1}{b_1} a_2 < b_2$$ but $$b_1 < a_1$$ so $$\frac {a_1}{b_1}<1$$ and $$a_2 < \frac {a_1}{b_1} a_2 < b_2$$.

.....

So we have $$a_1 >b_1 > b_2 >a_2$$.