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If $a_i$ and $b_i$ are positive, and $b= \sum ^n_{i=1} b_i$,$a= \sum ^n_{i=1} a_i$ prove $$\frac {ab}{a+b} \geq \sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i}$$ Additional: we should just use Cauchy inequality. However , if it's not possible, solve it with anything you want.

Things I have done so far:

Writing in different form $\frac {(\sum ^n_{i=1} b_i)(\sum ^n_{i=1} a_i)}{(\sum ^n_{i=1} b_i)+(\sum ^n_{i=1} a_i)} \geq \sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i}$

And maybe this could take me somewhere: ${(\sum ^n_{i=1} b_i)(\sum ^n_{i=1} a_i)} \geq (\sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i})({(\sum ^n_{i=1} b_i)+(\sum ^n_{i=1} a_i)})$

And another thing which came to my mind:

${(\sum ^n_{i=1} b_i)(\sum ^n_{i=1} a_i)} \geq (\sum ^n_{i=1} \sqrt {a_ib_i})^2$

$(\sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i})(\sum ^n_{i=1} a_i+b_i)\geq (\sum ^n_{i=1} \sqrt {a_ib_i})^2$

My main problem is find two parentheses to multiply them for writing Cauchy. I want some hint on this just to start working with inequality.

Update

It seems like no one can come with answer using Cauchy and without induction and Harmonic , Geometric and ... means.so if no answer come till tomorrow,I will give up and accept Liu Gang answer.

And I would appreciate someone could explain why it can't be solved using Cauchy and without induction and Harmonic , Geometric and ... means.(asking this because possibility of putting bounty)

2nd-Update

we can use any mean inequalities.but for only two number like $x$ and $y$.not generalized form for $n$ numbers.

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    $\begingroup$ An observation which may be helpful for intuition: if you multiply both sides by two, then one may write this inequality as $H(a,b)\geq \sum_i H(a_i,b_i)$ where $H(x,y)$ is the harmonic mean of $x$ and $y$. $\endgroup$ – Semiclassical Jul 29 '14 at 14:53
  • $\begingroup$ @Semiclassical,thanks for your hint.however,the teacher has yet to teach us means. so i'm not allowed to use them too. $\endgroup$ – user2838619 Jul 29 '14 at 14:56
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    $\begingroup$ to your 2nd-update: we know $xy \geq 0$ when they have the same sign. Thus if $x_1 \geq x_2$ if and only if $y_1 \geq y_2 $, then $(x_1 - x_2)(y_1 - y_2) \geq 0$, i.e. $x_1y_1 + x_2y_2 \geq x_1y_2 + x_2y_1$. Just to avoid applying the general rearrangement inequality $\endgroup$ – Petite Etincelle Jul 30 '14 at 9:10
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    $\begingroup$ I wonder if the rearrangement inequality for two terms is equivalent to the Cauchy-Schwarz inequality for two terms. $\endgroup$ – Jonas Dahlbæk Jul 30 '14 at 9:57
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Hint:

We only need to prove this is true for $n=2$. Since if it's ok for $n=2$, then

\begin{align} \frac{ab}{a+b} &= \frac{(a_1 + \sum_{i=2}^na_i)(b_1 + \sum_{i=2}^nb_i) }{a_1 + \sum_{i=2}^na_i + b_1 + \sum_{i=2}^nb_i} \\ &\geq \frac{a_1b_1}{a_1 + b_1} + \frac{(\sum_{i=2}^na_i)(\sum_{i=2}^nb_i)}{\sum_{i=2}^na_i + \sum_{i=2}^nb_i} \\ & \geq \cdots \geq \sum_{i=1}^n\frac{a_ib_i}{a_i + b_i} \end{align}

When $n=2$, the inequality is equivalent to \begin{align} \frac{a_1b_1}{a_1 + b_1}(a_2 + b_2) + \frac{a_2b_2}{a_2 + b_2}(a_1 + b_1) \leq a_1b_2 + a_2b_1 \end{align}

which can be proven by noting $x_1 = \frac{a_1}{a_1 + b_1}(a_2 + b_2)$, $y_2 = b_1$, $x_2 = a_2$, $y_1 = \frac{b_2}{a_2 + b_2}(a_1 + b_1)$ and using rearrangement inequality

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  • $\begingroup$ thanks for hint.I have two question.first did you used Cauchy?i can't find it by looking at inequality you write.and is this a proof by induction? $\endgroup$ – user2838619 Jul 29 '14 at 16:23
  • $\begingroup$ @user2838619 No, I didn't use Cauchy, I just assume this ineqaulity is true for $n=2$, then use this fact to deduce the inequality for any $n$. You can see it as a proof by induction. Still you need to prove that for $n=2$ this inequality is true $\endgroup$ – Petite Etincelle Jul 29 '14 at 16:37
  • $\begingroup$ @user2838619 I added the proof for $n=2$, but I didn't use Cauchy $\endgroup$ – Petite Etincelle Jul 29 '14 at 16:58
  • $\begingroup$ If I understand correctly, $x_1\geq x_2$ if and only if $y_1\geq y_2$, and that is why you can use the rearrangement inequality. What made you pick $x_j,y_j$ in the specific way you did? $\endgroup$ – Jonas Dahlbæk Jul 29 '14 at 18:27
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    $\begingroup$ @user161825 yeah exactly as what you said. I noticed $\frac{a_2 +b_2}{a_1 +b_1}$ and $\frac{a_1 +b_1}{a_2 +b_2}$ could make each other disappear and I wanted to make that happen and to get RHS at the same time, so... $\endgroup$ – Petite Etincelle Jul 29 '14 at 18:34
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This is really just a brute-force variation on the answer by Liu Gang. Consider the equivalent form $$ \sum_{i,j}\frac{a_ib_i(a_j+b_j)}{a_i+b_i}\leq\sum_{i,j}a_ib_j. $$ In these two sums, the diagonal terms with $i=j$ cancel, so we can equivalently show $$ \sum_{i\neq j}\frac{a_ib_i(a_j+b_j)}{a_i+b_i}\leq\sum_{i\neq j}a_ib_j. $$ In order to show this inequality, it is sufficient to show, for all $i\neq j$, that $$ \frac{a_ib_i(a_j+b_j)}{a_i+b_i}+\frac{a_jb_j(a_i+b_i)}{a_j+b_j}\leq a_ib_j+a_jb_i. $$ In other words, we have reduced to the case $n=2$. Let us, for notational convenience, put $i=1,j=2$. Here we could employ the rearrangement inequality to finish the argument. Let us instead multiply by $(a_1+b_1)(a_2+b_2)$ and cancel terms putting this inequality in the equivalent form $$ a_1b_1a_2b_2+a_2b_2a_1b_1\leq a_1^2b_2^2+a_2^2b_1^2. $$ Again, we could finish the argument by using the rearrangement inequality. Let us instead add $a_1^2b_1^2+a_2^2b_2^2$ to both sides, yeilding $$ \sum_{i,j=1}^2 a_i b_i a_j b_j\leq \sum_{i,j=1}^2 a_i^2b_j^2. $$ But this is the Cauchy-Schwarz inequality applied to the vectors $a=(a_1,a_2),b=(b_1,b_2)$ (which is, of course, a consequence of the rearrangement inequality).

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  • $\begingroup$ thanks for solution.here is another case which i noticed rearrangement inequality is more efficient(Andre answer).i'm unsure between accepting which answer,as Liu was first to answer and you used his base idea but you some how managed to use Cauchy. $\endgroup$ – user2838619 Jul 30 '14 at 11:09
  • $\begingroup$ I would consider my answer an addendum to Liu's answer, so please go ahead and accept his answer. $\endgroup$ – Jonas Dahlbæk Jul 30 '14 at 11:20

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