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 $a_1 > b_1 > b_2 > a_2$.

Question

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?

Answer 1

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 $$f(x)<g(x)$$ for every $x>0$. Hence $$(b_1-a_1)(b_1-a_2)=f(b_1)<g(b_1)=0$$ and $$(b_2-a_1)(b_2-a_2)=f(b_2)<g(b_2)=0.$$ Using $a_1>a_2$ and $b_1>b_2$, the two inequalities above show that $a_1>b_1>b_2>a_2$.

Answer 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 $$ a_1 > b_1 > b_2 > a_2. $$

Answer 3

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 $a_1 >b_1 > b_2 >a_2$.