Prove that: $$ \frac{a^2}{a+b} + \frac{b^2}{b+c} \geq \frac{3a+2b-c}{4} : (a, b, c)\in \mathbb{R}^+$$

This is just one of these questions where you just have no idea how to start. First impressions, I don't see how any known inequality can be used, and I also don't want to go just make everything as a sum then solve it.

I always was bad at inequalities and I don't know why. I did the other exercise just fine, but inequalities are hard for me. This is from a high-school olympiad.


$$\begin{align} \frac{a^2}{a+b} + \frac{b^2}{b+c} &\geqslant \frac{3a + 2b - c}{4}\\ \iff \frac{a^2}{a+b} - a + \frac{b^2}{b+c} - b &\geqslant - \frac{a + 2b + c}{4}\\ \iff -\frac{ab}{a+b} - \frac{bc}{b+c} &\geqslant - \frac{a+b}{4} - \frac{b+c}{4}\\ \iff \frac{(a+b)^2 - 4ab}{4(a+b)} + \frac{(b+c)^2 - 4bc}{4(b+c)} &\geqslant 0. \end{align}$$

Since all numbers are positive, the denominators are positive, and $(x+y)^2 - 4xy = (x-y)^2 \geqslant 0$.


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