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This question already has an answer here:

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

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marked as duplicate by Martin R, Surb, Macavity inequality Mar 5 '18 at 12:36

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  • $\begingroup$ great and now you can finish by induction no? $\endgroup$ – Surb Mar 5 '18 at 10:59
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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 $B_{n+1}=B_n+b_{n+1}$.

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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!

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