Inequality: $\frac{a+b+c}{2} \geq \frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}$ I am asking this question on behalf of another student, who sought my help through a school mentoring scheme, and claims that the question is similar to that found in Question $1$ of the British Maths Olympiad.
Prove that for all positive real numbers $a, b, c$:
a) $(a+b)^2 \geq 4ab$
b) $\frac{a+b+c}{2} \geq \frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}$
The person mentioned solved a) easily but struggled with b). To solve b), the person had attempted the following:
"...putting the right hand side as a single fraction and then cross multiplying as well as cancelling terms. But it turns out to be a repeated process which makes the original question even more complicated."


Looking at part a), I think the intention of part b) is
\begin{eqnarray*}\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}
& \stackrel{a)}{\leq} & \frac{(a+b)^2}{4(a+b)} +\frac{(b+c)^2}{4(b+c)} + \frac{(a+c)^2}{4(a+c)} \\
& = & \frac 14(a+b + b+c + a+c) \\
& = & \frac{a+b+c}{2}
\end{eqnarray*}
