# Prove inequality $\sum_{k=1}^{n}\frac{a_kb_k}{a_k+b_k}\le \frac{AB}{A+B}$ [duplicate]

Let $a_k, b_k (k=1,2,...,n)$ are real positive numbers. Prove the inequality $$\frac{a_1b_1}{a_1+b_1}+\frac{a_2b_2}{a_2+b_2}+...+\frac{a_n b_n}{a_n+b_n}\le \frac{AB} {A+B}$$ where $A=a_1+a_2+...+a_n$ and $B=b_1+b_2+...+b_n$

### My work so far:

If $n=2$ $$\frac{a_1b_1}{a_1+b_1}+\frac{a_2b_2}{a_2+b_2}\le \frac{(a_1+a_2)(b_1+b_2)}{a_1+a_2+b_1+b_2} \Leftrightarrow$$ $$\Leftrightarrow (a_1b_2-a_2b_1)^2\ge0$$

## marked as duplicate by Martin R, Surb, Macavity inequality StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 5 '18 at 12:36

Well, you're almost done. With the case $n=2$ you have proved, finish by induction: If, $$\sum_{i=1}^n \frac{a_ib_i}{a_i+b_i} \leq \frac{A_nB_n}{A_n+B_n}$$ then, $A_n>0$ and $B_n>0$ so using the case $n=2$, we get $$\sum_{i=1}^{n+1} \frac{a_ib_i}{a_i+b_i} \leq \frac{A_nB_n}{A_n+B_n}+\frac{a_{n+1}b_{n+1}}{a_{n+1}+b_{n+1}}\leq \frac{A_{n+1}B_{n+1}}{A_{n+1}+B_{n+1}}$$ with $A_{n+1}=A_n+a_{n+1}$ and $B_{n+1}=B_n+b_{n+1}$.
We need to prove that $$\sum_{i=1}^n\left(\frac{a_ib_i}{a_i+b_i}-\frac{a_i+b_i}{4}\right)\leq\frac{AB}{A+B}-\frac{A+B}{4}$$ or $$\sum_{k=1}^n\frac{(a_i-b_i)^2}{a_i+b_i}\geq\frac{(A-B)^2}{A+B},$$ which is C-S: $$\sum_{k=1}^n\frac{(a_i-b_i)^2}{a_i+b_i}\geq\frac{\left(\sum\limits_{i=1}^n(a_i-b_i)\right)^2}{\sum\limits_{k=1}^n(a_i+b_i)}=\frac{(A-B)^2}{A+B}.$$ Done!