Prove that $\frac{a_1^2}{a_1+b_1}+\cdots+\frac{a_n^2}{a_n+b_n} \geq \frac{1}{2}(a_1+\cdots+a_n).$

Question

Let $a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n$ be positive numbers with $a_1+a_2+\cdots+a_n = b_1+b_2+\cdots+b_n$. $$\text{Prove that} \dfrac{a_1^2}{a_1+b_1}+\cdots+\dfrac{a_n^2}{a_n+b_n} \geq \dfrac{1}{2}(a_1+\cdots+a_n).$$

Attempt

It seems like I should use AM-GM on the bottom of each fraction. We then get $\dfrac{a_i^2}{a_i+b_i} \leq \dfrac{a_i^2}{2\sqrt{a_ib_i}}$. But this doesn't seem to help as we get an upper bound. Since there is so much about $a_1+\cdots+a_n$ in this problem, I think a substitution for that might work.

Answer 1

Since $\displaystyle\sum_{i=1}^n\frac{a_i^{2}}{a_i+b_i}-\sum_{i=1}^{n}\frac{b_i^{2}}{a_i+b_i}=\sum_{i=1}^n\frac{a_i^{2}-b_i^{2}}{a_i+b_i}=\sum_{i=1}^n(a_i-b_i)=0$, $\displaystyle\;\;\sum_{i=1}^n\frac{a_i^{2}+b_i^{2}}{a_i+b_i}=2\sum_{i=1}^n\frac{a_i^{2}}{a_i+b_i}$.

Since $2(a_i^{2}+b_i^{2})\ge(a_i+b_i)^2$, $\displaystyle\;\;2\sum_{i=1}^n\frac{a_i^{2}}{a_i+b_i}=\sum_{i=1}^n\frac{a_i^{2}+b_i^{2}}{a_i+b_i}\ge\sum_{i=1}^n\frac{a_i+b_i}{2}=\sum_{i=1}^n a_i$

Answer 2

Using Cauchy-Schwarz inequality $$ (x_1^2+\cdots+x_n^2)(y_1^2+\cdots+y_n^2)\ge (x_1y_1+\cdots+x_ny_n)^2, $$ for $x_i=\displaystyle\frac{a_i}{\sqrt{a_i+b_i}}\,$ and $\,y_i=\sqrt{a_i+b_i}$, we obtain

$$ \left({\frac{a_1^2}{a_1+b_1}}+\cdots+\frac{a_n^2}{a_n+b_n}\right)\big((a_1+b_1)+\cdots+(a_n+b_n)\big)\ge \left(a_1+\cdots+a_n\right)^2, $$ or equivalently $$ 2\left({\frac{a_1^2}{a_1+b_1}}+\cdots+\frac{a_n^2}{a_n+b_n}\right)(a_1+\cdots+a_n)\ge \left(a_1+\cdots+a_n\right)^2, $$ and finally $$ {\frac{a_1^2}{a_1+b_1}}+\cdots+\frac{a_n^2}{a_n+b_n}\ge \frac{1}{2}\left(a_1+\cdots+a_n\right). $$

Answer 3

It can be proved by using the Cauchy reverse technique. In details, $$ \sum\limits_{i=1}^n{\frac{2a_i^2}{a_i+b_i}}=\sum\limits_{i=1}^n{\left(2a_i-\frac{2a_ib_i}{a_i+b_i}\right)} \ge \sum\limits_{i=1}^n{\left(2a_i-\sqrt{a_ib_i}\right)} $$ So it suffices to show the following inequality $$ \sum\limits_{i=1}^n{a_i} \ge \sum\limits_{i=1}^n{\sqrt{a_ib_i}} $$ , which is quite clear since $$ 2\sum\limits_{i=1}^n{a_i}=\sum\limits_{i=1}^n{\left(a_i+b_i\right)} \ge 2\sum\limits_{i=1}^n{\sqrt{a_ib_i}} $$ according to AM-GM inequality and the condition.

Remark. If there is a minus before a fraction, then we can apply AM-GM in the denominator.

Answer 4

The following solution is similar to the previous ones. First, we will prove the following lemma:

Lemma. For positive reals $x$ and $y$ the inequality $\frac{x^2}{y}\geq 2x-y$ holds.

Proof. Indeed, it's equivalent to $\frac{(x-y)^2}{y}\geq 0$.

Now, apply this inequality for $x=2a_i$ and $y=a_i+b_i$ $$ \frac{4a_i^2}{a_i+b_i}\geq 4a_i-(a_i+b_i)=3a_i-b_i. $$ Summing this inequalities for $i=\overline{1,n}$ we obtain $$ \sum_{i=1}^{n}\frac{a_i^2}{a_i+b_i}\geq\frac{1}{4}\sum_{i=1}^{n}(3a_i-b_i)=\frac{1}{2}\sum_{i=1}^{n}a_i, $$ as desired.

Comment. One can extend lemma in the following way: for all positive $x$, $y$, $p$ and $q$ (with $p\neq q$) we have $$ \frac{x^p}{y^q}\geq\frac{px^{p-q}-qy^{p-q}}{p-q}. $$ Using this inequality we can create a lot of similar problems (this lemma helps when we want to estimate fraction of type $\frac{A^p}{B^q}$).

Answer 5

Hint: Use the fact that $f(x)=\frac{1}{x+1}$ is convex and then use weighted Jensen's inequality .