This one easy By Caushy Swarz Inequality, Am-Gm,... algebraic substitution all of them kill this problem. Applying C-S : (LHS)(Sum a_n+Sum b_n)>= (Sum a_n)^2 so our problem is killed if you want i can do that by Am-Gm too Davron

$\sum \frac{a_i^2}{a_i+b_i} \ge \frac{\left(\sum a_i\right)^2}{\sum a_i+ \sum b_i}$ by Cauchy, but since $\sum a_i+\sum b_i = 2 \sum a_i$, we have $\frac{\left(\sum a_i\right)^2}{\sum a_i+ \sum b_i} = \frac{\left(\sum a_i\right)^2}{2\sum a_i} = \frac{\sum a_i}{2}$.

My proof is based on only AM-GM. \[ \frac{a^2_{i}}{a_{i}+b_{i}} + \frac{a_{i}+b_{i}}{4} \ge a_{i}\] Therefore, $\sum \frac{a^2_{i}}{a_{i}+b_{i}} + \sum \frac{a_{i}}{2} \ge \sum a_{i}$. so we get our result.

This is just Titu's Lemma applied once

shobber wrote: Let $a_1$, $a_2$, $\cdots$, $a_n$, $b_1$, $b_2$, $\cdots$, $b_n$ be positive real numbers such that $a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$. Show that \[ \frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2} \] Let $f(x) = \frac{1}{1+x}$. f is convex and we have: \[\frac{1}{\sum a_{i}}\sum \frac{a_{i}^2}{a_{i}+b_{i}}=\sum \frac{a_{i}}{\sum a_{i}}f(\frac{b_{i}}{a_{i}}) \geq f(\frac{\sum b_{i}}{\sum a_{i}})=f(1)=\frac{1}{2}\]

Yet another proof: In what follows $\sum $ denotes $\sum_{i=1}^{n}$. Let $A=\sum \frac{a_i^2}{a_i+b_i}$ and $B=\sum \frac{b_i^2}{a_i+b_i}$. Note that $$A-B=\sum\frac{a_i^2-b_i^2}{a_i+b_i}=\sum a_i-b_i=\sum a_i-\sum b_i=0\implies A=B$$Also note that for all $i$, $\frac{a_i^2+b_i^2}{a_i+b_i}\ge \frac{a_i+b_i}{2}\iff \left(a_i-b_i\right)^2\ge 0$, which is true. Thus $$A=\frac 12 (A+B)\ge \frac 12 \sum \frac{a_i^2+b_i^2}{a_i+b_i}\ge \frac 12 \sum \frac{a_i+b_i}{2}=\frac 12 \left(\sum a_i\right)$$QED.

shobber wrote: Let $a_1$, $a_2$, $\cdots$, $a_n$, $b_1$, $b_2$, $\cdots$, $b_n$ be positive real numbers such that $a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$. Show that \[ \frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2} \]

Solution. By Titu's Lemma we have that $\frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{(a_1+a_2+\cdots +a_n)^2}{a_1+a_2+\cdots +a_n+b_1+b_2+\cdots +b_n}$, $a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n \implies$ $\frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2}.\blacksquare$

this problem can be easily solved by using Titu's Lemma By titu's lemma we can say that, $\sum \limits_{i=1}^{n}$$\frac{a_i^2}{a_i+b_i}\geq \frac{(a_1+a_2+a_3+ \cdots +a_n)^2}{a_1+b_1+a_2+b_2+ \cdots +a_n+b_n}$ $As,a_1+a_2+a_3+ \cdots +a_n=b_1+b_2+b_3+ \cdots +b_n$ (stated in the problem) So,$a_1+b_1+a_2+b_2+ \cdots +a_n+b_n=2(a_1+a_2+ \cdots +a_n)$ Now,we can easily say that, $\sum \limits_{i=1}^{n}$$\frac{a_i^2}{a_i+b_i}\geq \frac{(a_1+a_2+\cdots +a_n)^2}{2(a_1+b_1+a_2+b_2+ \cdots +a_n+ b_n}$ $\Rightarrow \sum \limits_{i=1}^{n}\frac{a_i^2}{a_i+b_i} \geq \frac {a_1+a_2+ \cdots a_n}{2}$

This problem is a direct application of titu lemma

shobber wrote: Let $a_1$, $a_2$, $\cdots$, $a_n$, $b_1$, $b_2$, $\cdots$, $b_n$ be positive real numbers such that $a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$. Show that \[ \frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2} \] Direct titu

Titu's Lemma gives \[ \frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{(a_1+a_2+\dots+a_n)^2}{a_1+a_2+\dots+a_n+b_1+b_2+\dots+b_n}.\]But $a_1+a_2+\dots+a_n=b_1+b_2+\dots+b_n$ implies that \[\frac{(a_1+a_2+\dots+a_n)^2}{a_1+a_2+\dots+a_n+b_1+b_2+\dots+b_n}=\frac{(a_1+a_2+\dots+a_n)^2}{2(a_1+a_2+\dots+a_n)}=\frac{a_1+a_2+\dots+a_n}{2},\]which proves the desired inequality.

Let $A=\sum_{k=1}^na_k$ and $B=\sum_{k=1}^nb_k$. We have $A=B$. $$\sum_{k=1}^n\frac{a_k^2}{a_k+b_k}\overset{\text{T2}}\ge\frac{A^2}{A+B}=\frac A2.\square$$

This does not even require Cauchy-Schwarz or Engel form. Just $$\dfrac{a+b}{2}\geq \dfrac{2ab}{a+b}$$is enough to prove this. It's easy to forget that Engel form was not as well known in 1991. See the following \begin{align*} \dfrac{a_{1}^{2}}{a_{1}+b_{1}}+\ldots+\dfrac{a_{n}^{2}}{a_{n}+b_{n}} & \geq\dfrac{a_{1}+\ldots+a_{n}}{2}\\ \iff \dfrac{a_{1}^{2}}{a_{1}+b_{1}}+\ldots+\dfrac{a_{n}^{2}}{a_{n}+b_{n}}+\dfrac{a_{1}+\ldots+a_{n}}{2} & \geq a_{1}+\ldots+a_{n}\\ \iff \dfrac{a_{1}+\ldots+a_{n}}{2} & \geq \left(a_{1}-\dfrac{a_{1}^{2}}{a_{1}+b_{1}}\right)+\ldots+\left(a_{n}-\dfrac{a_{n}^{2}}{a_{n}+b_{n}}\right)\\ & = \dfrac{a_{1}b_{1}}{a_{1}+b_{1}}+\ldots+\dfrac{a_{n}b_{n}}{a_{n}+b_{n}}\\ \iff a_{1}+\ldots+a_{n} & \geq \dfrac{2a_{1}b_{1}}{a_{1}+b_{1}}+\ldots+\dfrac{2a_{n}b_{n}}{a_{n}+b_{n}}\\ \iff \dfrac{1}{2}(a_{1}+b_{1}+\ldots+a_{n}+b_{n}) & \geq \dfrac{2a_{1}b_{1}}{a_{1}+b_{1}}+\ldots+\dfrac{2a_{n}b_{n}}{a_{n}+b_{n}}\\ \iff \dfrac{a_{1}+b_{1}}{2}+\ldots+\dfrac{a_{n}+b_{n}}{2} & \geq \dfrac{2a_{1}b_{1}}{a_{1}+b_{1}}+\ldots+\dfrac{2a_{n}b_{n}}{a_{n}+b_{n}} \end{align*}The last inequality is evidently true.

Hello, I am new to Olympiad math . In very likely case that this is wrong, I would appreciate some feedback on what my mistake is. $a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$ wouldn't this mean that $a_n= b_n$? Therefore, \[ \frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2} \] Can be also written as \[ \frac{a_1^2}{2(a_1)} + \frac{a_2^2}{2(a_2)} + \cdots + \frac{a_n^2}{2(a_n)} \geq \frac{a_1 + a_2 + \cdots + a_n}{2} \] When both sides are multiplied by 2, we get that $a_1 + a_2 + \cdots + a_n \geq $$a_1 + a_2 + \cdots + a_n$ I think this is a viable proof. Please correct me if there are mistakes

@above, $$a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$$doesn't mean that $a_n= b_n$, for example $$1 + 2 + 3 = 1 + 1 + 4$$

I presumed that there would be some kind of formula behind the a an b values. Are you saying that the a and b values can be from sets with no pattern?

@above Yeah the only condition ${a_i}$ and ${b_i}$ must satisfy is that they are positive real numbers and, $$a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$$ other than that they need not have any pattern. You can refer to previous posts for the correct solution

Alright thanks

1990 USSR or 1990 All-Russian https://artofproblemsolving.com/community/u229790h1895500p27134878

Another: $$\frac{a^2}{b} \ge 2a-b, \text{ because } (a-b)^2 \ge 0$$. Now inequality is obviously.

$\color{red} \boxed{\textbf{SOLUTION}}$ From the $\textbf{Engel Form of Cauchy-Schwarz Inequality,}$ $$\sum \frac{a_i^2}{a_i+b_i} \ge \frac{\left(\sum a_i\right)^2}{\sum a_i+ \sum b_i} = \frac{\left(\sum a_i\right)^2}{2\sum a_i} = \frac{\sum a_i}{2}\blacksquare$$

WHOA!! DO WHATEVER YOU WANT ...TITU,C-S, WHATEVER

One line Titu's Lemma

Observe that from Titu's lemma, $$\sum_{i = 1}^n \frac{a_i^2}{a_i+b_i} \ge \frac{\left( \sum_{i = 1}^n a_i\right)^2}{\sum_{i=1}^n a_i + b_i} = \frac{\left( \sum_{i = 1}^n a_i\right)^2}{2\sum_{i=1}^n a_i} = \frac{\sum_{i=1}^n a_i}{2},$$qed.$\square$

Observe that \[\sum \frac{a_i^2}{a_i+b_i} \ge \frac{(\sum a_i)^2}{\sum (a_i+b_i)} = \frac{(\sum a_i)^2}{2\sum a_i} = \frac{a_1 + a_2 + \dots + a_n}{2},\] as desired. $\square$

$$\frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{(a_1 + a_2 + \cdots + a_n)^2}{a_1 + a_2 + \cdots + a_n + b_1 + b_2 + \cdots + b_n} = \frac{(a_1 + a_2 + \cdots + a_n)^2}{2(a_1 + a_2 + \cdots + a_n)} = \frac{a_1 + a_2 + \cdots + a_n}{2} $$