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Algebra Difficulty 5.2 AIME, harder Prove it Saudi Arabia

Let a>0a > 0 be a real number and let f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} be a function satisfying
f(x1)+f(x2)≥af(x1+x2),∀x1,x2∈R. f\left(x_{1}\right) + f\left(x_{2}\right) \geq a f\left(x_{1} + x_{2}\right), \forall x_{1}, x_{2} \in \mathbb{R} .
Prove that
f(x1)+f(x2)+f(x3)≥3a2a+2f(x1+x2+x3),∀x1,x2,x3∈R. f\left(x_{1}\right) + f\left(x_{2}\right) + f\left(x_{3}\right) \geq \frac{3 a^{2}}{a+2} f\left(x_{1} + x_{2} + x_{3}\right), \forall x_{1}, x_{2}, x_{3} \in \mathbb{R}.

Solution

Using the inequality in the hypothesis we get successively:
f(x1)+f(x2)+af(x3)≥a2f(x1+x2+x3)f(x1)+af(x2)+f(x3)≥a2f(x1+x2+x3)af(x1)+f(x2)+f(x3)≥a2f(x1+x2+x3) \begin{aligned} & f\left(x_{1}\right) + f\left(x_{2}\right) + a f\left(x_{3}\right) \geq a^{2} f\left(x_{1} + x_{2} + x_{3}\right) \\ & f\left(x_{1}\right) + a f\left(x_{2}\right) + f\left(x_{3}\right) \geq a^{2} f\left(x_{1} + x_{2} + x_{3}\right) \\ & a f\left(x_{1}\right) + f\left(x_{2}\right) + f\left(x_{3}\right) \geq a^{2} f\left(x_{1} + x_{2} + x_{3}\right) \end{aligned}
Add all and get:
(a+2)(f(x1)+f(x2)+f(x3))≥3a2f(x1+x2+x3) (a+2)\left(f\left(x_{1}\right) + f\left(x_{2}\right) + f\left(x_{3}\right)\right) \geq 3 a^{2} f\left(x_{1} + x_{2} + x_{3}\right)
and the conclusion follows.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.